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Factoring the polynomial $5c^2 - 52c + 20$
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I was trying to find the factors of that expression above:
$5c^2-52c+20.$
The solution is
$(5c−2)(c−10).$
I don't understand: how did we get these values $-2$ and $-10?$ Is there any way to solve it quickly by looking at some parts of that expression?
algebra-precalculus factoring
edited Aug 2 at 9:25
N. F. Taussig
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asked Aug 1 at 17:41
user10133549
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I was trying to find the factors of that expression above:
$5c^2-52c+20.$
The solution is
$(5c−2)(c−10).$
I don't understand: how did we get these values $-2$ and $-10?$ Is there any way to solve it quickly by looking at some parts of that expression?
algebra-precalculus factoring
edited Aug 2 at 9:25
N. F. Taussig
38k93053
asked Aug 1 at 17:41
user10133549
61
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was trying to find the factors of that expression above:
$5c^2-52c+20.$
The solution is
$(5c−2)(c−10).$
I don't understand: how did we get these values $-2$ and $-10?$ Is there any way to solve it quickly by looking at some parts of that expression?
algebra-precalculus factoring
edited Aug 2 at 9:25
N. F. Taussig
38k93053
asked Aug 1 at 17:41
user10133549
61
I was trying to find the factors of that expression above:
$5c^2-52c+20.$
The solution is
$(5c−2)(c−10).$
I don't understand: how did we get these values $-2$ and $-10?$ Is there any way to solve it quickly by looking at some parts of that expression?
algebra-precalculus factoring
edited Aug 2 at 9:25
N. F. Taussig
38k93053
asked Aug 1 at 17:41
user10133549
61
edited Aug 2 at 9:25
N. F. Taussig
38k93053
edited Aug 2 at 9:25
N. F. Taussig
38k93053
edited Aug 2 at 9:25
N. F. Taussig
38k93053
38k93053
asked Aug 1 at 17:41
user10133549
61
asked Aug 1 at 17:41
user10133549
61
asked Aug 1 at 17:41
user10133549
61
61
add a comment |Â
add a comment |Â
6 Answers
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$$5c^2-52c+20=0$$
$$5c^2-50c-2c+20=0\5c(c-10)-2(c-10)=0$$
$$(5c-2)(c-10)=0$$
or you can use quadratic formula $$x=dfrac-bpm sqrtb^2-4ac2a$$
Here $a=5,b=-52,c=20$
answered Aug 1 at 17:44
Key Flex
3,797422
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
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Break the expression into the groups
$$left(5c^2-50cright)+left(-2c+20right)$$
Factor out $5c$ from $5c^2-50$, which yields $5cleft(c-10right)$. Now factor out $-2$ from $-2c+20$ and we get $-2left(c-10right)$
Factor out the common term $left(c-10right)$
$$left(c-10right)left(5c-2right)$$
answered Aug 1 at 17:47
C.Maxwell
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There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):
- First write $left(dfracaxphantom+4phantom5right)
left(dfracaxphantom+4phantom5right).$ - Find the product $ac$, including sign.
- Find the prime factorization of $ac$ using the factor tree.
- Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $sqrtac$.
- Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.
- Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$
- Divide each of these binomials by its own GCF:
$left(dfracax+soperatornamegcf(a,s)right)
left(dfracax+toperatornamegcf(a,t)right).$ Check that
$$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$
answered Aug 1 at 17:50
Adrian Keister
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Consider what happens when you multiply two linear polynomials.
beginalign*
(4x - 5)(3x + 8) & = 4x(3x + 8) - 5(3x + 8) && textapply the distributive law\
& = colorblue12x^2 + colorgreen32x - colorgreen15x - colorblue40 && textapply the distributive law\
& = colorblue12x^2 + colorgreen17x - colorblue40 && textcombine like terms
endalign*
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients that combine to form the linear coefficient of the product, that is,
$$(colorblue12)(colorblue-40) = (colorgreen32)( colorgreen-15)$$
To factor $12x^2 + 17x - 40$, we wish to carry out these steps in reverse.
beginalign*
12x^2 + colorgreen17x - 40 & = 12x^2 + 32x - 15x - 40 && textsplit the linear term\
& = 4x(3x + 8) - 5(3x + 8) && textfactor by grouping\
& = (4x - 5)(3x + 8) && textextract the common factor
endalign*
The observation we made above enables us to split the linear term. To do so, we must find two numbers with product $(colorblue12)(colorblue-40) = -480$ that have sum $colorgreen17$. We can do so by listing the factor pairs of $-480$, then finding the pair in which the sum of the factors is $colorgreen17$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
(1)(-480) & -479 & (-1)(480) & 479\
(2)(-240) & -238 & (-2)(240) & 238\
(3)(-160) & -157 & (-3)(160) & 157\
(4)(-120) & -116 & (-4)(120) & 116\
(5)(-96) & -91 & (-5)(96) & 91\
(6)(-80) & -74 & (-6)(80) & 74\
(8)(-60) & -52 & (-8)(60) & 52\
(10)(-48) & -38 & (-10)(48) & 38\
(12)(-40) & -28 & (-12)(40) & 28\
(15)(-32) & -17 & (colorgreen-15)(colorgreen32) & colorgreen17\
(20)(-24) & -4 & (-20)(24) & 4
endarray
The desired factors are $colorgreen32$ and $colorgreen-15$. Once we split the linear term, we extract the common factor from the first two terms and the common factor from the last two terms, which is called factoring by grouping. Finally, we extract the common linear factor to complete the factorization.
Let's apply this technique to your example.
$$5c^2 - 52c + 20$$
To split the linear term, we must find two numbers with product $colorblue5 cdot colorblue20 = 100$ and sum $colorgreen-52$. We list the factor pairs of $100$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
1 cdot 100 & 101 & (-1)(-100) & -101\
2 cdot 50 & 52 & (colorgreen-2)(colorgreen-50) & colorgreen-52\
4 cdot 25 & 29 & (-4)(-25) & -29\
5 cdot 20 & 25 & (-5)(-20) & -25\
10 cdot 10 & 20 & (-10)(-10) & -20
endarray
The desired factors are $colorgreen-2$ and $colorgreen-50$. Hence,
beginalign*
5c^2 - 52c + 20 & = 5c^2 - 2c - 50c + 20 && textsplit the linear term\
& = c(5c - 2) - 10(5c - 2) && textfactor by grouping\
& = (c - 10)(5c - 2) && textextract the common factor
endalign*
More generally, a factorization
$$ax^2 + bx + c = (rx + s)(tx + u)$$
of a quadratic polynomial $ax^2 + bx + c$ with respect to the rational numbers can be found if $a$, $b$, and $c$ are integers such that there exist integers with product $ac$ and sum $b$.
Caveat: Almost every quadratic polynomial you can write down is irreducible with respect to the rational numbers, meaning it cannot be split into two linear factors with rational coefficients. For instance, $x^2 + 2x + 3$ is irreducible with respect to the rational numbers since there are no integers with product $1 cdot 3 = 3$ and sum $2$.
answered Aug 2 at 17:50
N. F. Taussig
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There's always completing the square and/or the quadratic formula.
And bear in mind not all quadratic expression factor to rational components.
But:
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
Let's explain:
So if we solve $5c^2 - 52c + 20 = 0$ by completing the square (or quadratic formula) we get:
$5c^2 - 52c + 20 = 0$
$c^2 - frac 525c + 4 = 0$
$c^2 - frac 525c = -4$
$c^2 - frac 525c + (frac 265)^2 $
$(c - frac 265)^2 = = (frac 265)^2 - 4=frac 4*13^225 - frac 4*2525= frac 4(13^2 - 5^2)25 = frac 4*14425=(frac 2*125)^2$
$c -frac 265 = pm frac 245$
$c = frac 265 pm 245$
$c = frac 25$ or $c = 10$ so
.....
Now
Suppose $5c^2 - 52c + 20$ factored into $(mc + u)(nc + v)=mn(c +frac um)(c+frac vn)$. Then $5c^2 - 52c + 20 = 0 iff (c +frac um)(c+frac vn) = 0 iff c= -frac um$ or $c = -frac vn$.
So we know that the solutions are $c = frac 25$ and $c = 10$ so we must have that $5c^2 - 52c + 20$ factors to $5(c-frac 25)(c-10)$ or $(5c - 2)(c-20)$.
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
....
That's how you do it if you just can't see anything else.
In general you can make educated guesses:
If $5c^2 + 52c + 20 = (mc + u)(nc+ v)$ then
$mn = 5$
$un + vm = -52$
$uv = 20$.
If this behaves "nicely" (which you have no reason to assume it will!) then $mn = 5$ means $m,n = 5,1$.
And $uv = 20$ so $u,v =pm 1,20, pm 2,10, pm 4,5$
We have $u + 5v = -52$. The mean $u = -2, - 7 ,-12....$ and $v = -10, -9, -8....$ etc. The only choses we have that work are $-2,-10$.
So it was a very educated guess that $(5c -2)(c-10) =5c^2 + (-2-5*10)c +(-2*-10) = 5c^2 -52c + 20$.
answered Aug 2 at 18:52
fleablood
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$$beginalign
5c^2 - 52c + 20 &= frac55cdotbig(5c^2 - 52c + 20big)\
&= frac25c^2 - 260c + 1005\
&= frac(5c)^2 -52(5c) + 1005\
&= fract^2 -52t + 1005text where t=5c\
endalign$$
Can you take it from there?
answered 2 days ago
John Joy
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6 Answers
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6 Answers
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$$5c^2-52c+20=0$$
$$5c^2-50c-2c+20=0\5c(c-10)-2(c-10)=0$$
$$(5c-2)(c-10)=0$$
or you can use quadratic formula $$x=dfrac-bpm sqrtb^2-4ac2a$$
Here $a=5,b=-52,c=20$
answered Aug 1 at 17:44
Key Flex
3,797422
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
add a comment |Â
up vote
1
down vote
$$5c^2-52c+20=0$$
$$5c^2-50c-2c+20=0\5c(c-10)-2(c-10)=0$$
$$(5c-2)(c-10)=0$$
or you can use quadratic formula $$x=dfrac-bpm sqrtb^2-4ac2a$$
Here $a=5,b=-52,c=20$
answered Aug 1 at 17:44
Key Flex
3,797422
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$5c^2-52c+20=0$$
$$5c^2-50c-2c+20=0\5c(c-10)-2(c-10)=0$$
$$(5c-2)(c-10)=0$$
or you can use quadratic formula $$x=dfrac-bpm sqrtb^2-4ac2a$$
Here $a=5,b=-52,c=20$
answered Aug 1 at 17:44
Key Flex
3,797422
$$5c^2-52c+20=0$$
$$5c^2-50c-2c+20=0\5c(c-10)-2(c-10)=0$$
$$(5c-2)(c-10)=0$$
or you can use quadratic formula $$x=dfrac-bpm sqrtb^2-4ac2a$$
Here $a=5,b=-52,c=20$
answered Aug 1 at 17:44
Key Flex
3,797422
answered Aug 1 at 17:44
Key Flex
3,797422
answered Aug 1 at 17:44
Key Flex
3,797422
answered Aug 1 at 17:44
Key Flex
3,797422
3,797422
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
add a comment |Â
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
This is good but how would you answer if you student asked/answered "But we don't know that $5c^2 - 52c + 20 = 0$. We need to factor it for all possible answers."
– fleablood
Aug 2 at 18:54
add a comment |Â
up vote
0
down vote
Break the expression into the groups
$$left(5c^2-50cright)+left(-2c+20right)$$
Factor out $5c$ from $5c^2-50$, which yields $5cleft(c-10right)$. Now factor out $-2$ from $-2c+20$ and we get $-2left(c-10right)$
Factor out the common term $left(c-10right)$
$$left(c-10right)left(5c-2right)$$
answered Aug 1 at 17:47
C.Maxwell
131
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Break the expression into the groups
$$left(5c^2-50cright)+left(-2c+20right)$$
Factor out $5c$ from $5c^2-50$, which yields $5cleft(c-10right)$. Now factor out $-2$ from $-2c+20$ and we get $-2left(c-10right)$
Factor out the common term $left(c-10right)$
$$left(c-10right)left(5c-2right)$$
answered Aug 1 at 17:47
C.Maxwell
131
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up vote
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up vote
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down vote
Break the expression into the groups
$$left(5c^2-50cright)+left(-2c+20right)$$
Factor out $5c$ from $5c^2-50$, which yields $5cleft(c-10right)$. Now factor out $-2$ from $-2c+20$ and we get $-2left(c-10right)$
Factor out the common term $left(c-10right)$
$$left(c-10right)left(5c-2right)$$
answered Aug 1 at 17:47
C.Maxwell
131
Break the expression into the groups
$$left(5c^2-50cright)+left(-2c+20right)$$
Factor out $5c$ from $5c^2-50$, which yields $5cleft(c-10right)$. Now factor out $-2$ from $-2c+20$ and we get $-2left(c-10right)$
Factor out the common term $left(c-10right)$
$$left(c-10right)left(5c-2right)$$
answered Aug 1 at 17:47
C.Maxwell
131
answered Aug 1 at 17:47
C.Maxwell
131
answered Aug 1 at 17:47
C.Maxwell
131
answered Aug 1 at 17:47
C.Maxwell
131
131
add a comment |Â
add a comment |Â
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0
down vote
There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):
- First write $left(dfracaxphantom+4phantom5right)
left(dfracaxphantom+4phantom5right).$ - Find the product $ac$, including sign.
- Find the prime factorization of $ac$ using the factor tree.
- Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $sqrtac$.
- Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.
- Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$
- Divide each of these binomials by its own GCF:
$left(dfracax+soperatornamegcf(a,s)right)
left(dfracax+toperatornamegcf(a,t)right).$ Check that
$$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$
answered Aug 1 at 17:50
Adrian Keister
3,49321533
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up vote
0
down vote
There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):
- First write $left(dfracaxphantom+4phantom5right)
left(dfracaxphantom+4phantom5right).$ - Find the product $ac$, including sign.
- Find the prime factorization of $ac$ using the factor tree.
- Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $sqrtac$.
- Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.
- Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$
- Divide each of these binomials by its own GCF:
$left(dfracax+soperatornamegcf(a,s)right)
left(dfracax+toperatornamegcf(a,t)right).$ Check that
$$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$
answered Aug 1 at 17:50
Adrian Keister
3,49321533
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):
- First write $left(dfracaxphantom+4phantom5right)
left(dfracaxphantom+4phantom5right).$ - Find the product $ac$, including sign.
- Find the prime factorization of $ac$ using the factor tree.
- Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $sqrtac$.
- Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.
- Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$
- Divide each of these binomials by its own GCF:
$left(dfracax+soperatornamegcf(a,s)right)
left(dfracax+toperatornamegcf(a,t)right).$ Check that
$$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$
answered Aug 1 at 17:50
Adrian Keister
3,49321533
There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):
- First write $left(dfracaxphantom+4phantom5right)
left(dfracaxphantom+4phantom5right).$ - Find the product $ac$, including sign.
- Find the prime factorization of $ac$ using the factor tree.
- Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $sqrtac$.
- Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.
- Write $left(dfracax+sphantom5right)left(dfracax+tphantom5right).$
- Divide each of these binomials by its own GCF:
$left(dfracax+soperatornamegcf(a,s)right)
left(dfracax+toperatornamegcf(a,t)right).$ Check that
$$operatornamegcf(a,s)cdotoperatornamegcf(a,t)=a.$$
answered Aug 1 at 17:50
Adrian Keister
3,49321533
answered Aug 1 at 17:50
Adrian Keister
3,49321533
answered Aug 1 at 17:50
Adrian Keister
3,49321533
answered Aug 1 at 17:50
Adrian Keister
3,49321533
3,49321533
add a comment |Â
add a comment |Â
up vote
0
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Consider what happens when you multiply two linear polynomials.
beginalign*
(4x - 5)(3x + 8) & = 4x(3x + 8) - 5(3x + 8) && textapply the distributive law\
& = colorblue12x^2 + colorgreen32x - colorgreen15x - colorblue40 && textapply the distributive law\
& = colorblue12x^2 + colorgreen17x - colorblue40 && textcombine like terms
endalign*
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients that combine to form the linear coefficient of the product, that is,
$$(colorblue12)(colorblue-40) = (colorgreen32)( colorgreen-15)$$
To factor $12x^2 + 17x - 40$, we wish to carry out these steps in reverse.
beginalign*
12x^2 + colorgreen17x - 40 & = 12x^2 + 32x - 15x - 40 && textsplit the linear term\
& = 4x(3x + 8) - 5(3x + 8) && textfactor by grouping\
& = (4x - 5)(3x + 8) && textextract the common factor
endalign*
The observation we made above enables us to split the linear term. To do so, we must find two numbers with product $(colorblue12)(colorblue-40) = -480$ that have sum $colorgreen17$. We can do so by listing the factor pairs of $-480$, then finding the pair in which the sum of the factors is $colorgreen17$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
(1)(-480) & -479 & (-1)(480) & 479\
(2)(-240) & -238 & (-2)(240) & 238\
(3)(-160) & -157 & (-3)(160) & 157\
(4)(-120) & -116 & (-4)(120) & 116\
(5)(-96) & -91 & (-5)(96) & 91\
(6)(-80) & -74 & (-6)(80) & 74\
(8)(-60) & -52 & (-8)(60) & 52\
(10)(-48) & -38 & (-10)(48) & 38\
(12)(-40) & -28 & (-12)(40) & 28\
(15)(-32) & -17 & (colorgreen-15)(colorgreen32) & colorgreen17\
(20)(-24) & -4 & (-20)(24) & 4
endarray
The desired factors are $colorgreen32$ and $colorgreen-15$. Once we split the linear term, we extract the common factor from the first two terms and the common factor from the last two terms, which is called factoring by grouping. Finally, we extract the common linear factor to complete the factorization.
Let's apply this technique to your example.
$$5c^2 - 52c + 20$$
To split the linear term, we must find two numbers with product $colorblue5 cdot colorblue20 = 100$ and sum $colorgreen-52$. We list the factor pairs of $100$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
1 cdot 100 & 101 & (-1)(-100) & -101\
2 cdot 50 & 52 & (colorgreen-2)(colorgreen-50) & colorgreen-52\
4 cdot 25 & 29 & (-4)(-25) & -29\
5 cdot 20 & 25 & (-5)(-20) & -25\
10 cdot 10 & 20 & (-10)(-10) & -20
endarray
The desired factors are $colorgreen-2$ and $colorgreen-50$. Hence,
beginalign*
5c^2 - 52c + 20 & = 5c^2 - 2c - 50c + 20 && textsplit the linear term\
& = c(5c - 2) - 10(5c - 2) && textfactor by grouping\
& = (c - 10)(5c - 2) && textextract the common factor
endalign*
More generally, a factorization
$$ax^2 + bx + c = (rx + s)(tx + u)$$
of a quadratic polynomial $ax^2 + bx + c$ with respect to the rational numbers can be found if $a$, $b$, and $c$ are integers such that there exist integers with product $ac$ and sum $b$.
Caveat: Almost every quadratic polynomial you can write down is irreducible with respect to the rational numbers, meaning it cannot be split into two linear factors with rational coefficients. For instance, $x^2 + 2x + 3$ is irreducible with respect to the rational numbers since there are no integers with product $1 cdot 3 = 3$ and sum $2$.
answered Aug 2 at 17:50
N. F. Taussig
38k93053
add a comment |Â
up vote
0
down vote
Consider what happens when you multiply two linear polynomials.
beginalign*
(4x - 5)(3x + 8) & = 4x(3x + 8) - 5(3x + 8) && textapply the distributive law\
& = colorblue12x^2 + colorgreen32x - colorgreen15x - colorblue40 && textapply the distributive law\
& = colorblue12x^2 + colorgreen17x - colorblue40 && textcombine like terms
endalign*
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients that combine to form the linear coefficient of the product, that is,
$$(colorblue12)(colorblue-40) = (colorgreen32)( colorgreen-15)$$
To factor $12x^2 + 17x - 40$, we wish to carry out these steps in reverse.
beginalign*
12x^2 + colorgreen17x - 40 & = 12x^2 + 32x - 15x - 40 && textsplit the linear term\
& = 4x(3x + 8) - 5(3x + 8) && textfactor by grouping\
& = (4x - 5)(3x + 8) && textextract the common factor
endalign*
The observation we made above enables us to split the linear term. To do so, we must find two numbers with product $(colorblue12)(colorblue-40) = -480$ that have sum $colorgreen17$. We can do so by listing the factor pairs of $-480$, then finding the pair in which the sum of the factors is $colorgreen17$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
(1)(-480) & -479 & (-1)(480) & 479\
(2)(-240) & -238 & (-2)(240) & 238\
(3)(-160) & -157 & (-3)(160) & 157\
(4)(-120) & -116 & (-4)(120) & 116\
(5)(-96) & -91 & (-5)(96) & 91\
(6)(-80) & -74 & (-6)(80) & 74\
(8)(-60) & -52 & (-8)(60) & 52\
(10)(-48) & -38 & (-10)(48) & 38\
(12)(-40) & -28 & (-12)(40) & 28\
(15)(-32) & -17 & (colorgreen-15)(colorgreen32) & colorgreen17\
(20)(-24) & -4 & (-20)(24) & 4
endarray
The desired factors are $colorgreen32$ and $colorgreen-15$. Once we split the linear term, we extract the common factor from the first two terms and the common factor from the last two terms, which is called factoring by grouping. Finally, we extract the common linear factor to complete the factorization.
Let's apply this technique to your example.
$$5c^2 - 52c + 20$$
To split the linear term, we must find two numbers with product $colorblue5 cdot colorblue20 = 100$ and sum $colorgreen-52$. We list the factor pairs of $100$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
1 cdot 100 & 101 & (-1)(-100) & -101\
2 cdot 50 & 52 & (colorgreen-2)(colorgreen-50) & colorgreen-52\
4 cdot 25 & 29 & (-4)(-25) & -29\
5 cdot 20 & 25 & (-5)(-20) & -25\
10 cdot 10 & 20 & (-10)(-10) & -20
endarray
The desired factors are $colorgreen-2$ and $colorgreen-50$. Hence,
beginalign*
5c^2 - 52c + 20 & = 5c^2 - 2c - 50c + 20 && textsplit the linear term\
& = c(5c - 2) - 10(5c - 2) && textfactor by grouping\
& = (c - 10)(5c - 2) && textextract the common factor
endalign*
More generally, a factorization
$$ax^2 + bx + c = (rx + s)(tx + u)$$
of a quadratic polynomial $ax^2 + bx + c$ with respect to the rational numbers can be found if $a$, $b$, and $c$ are integers such that there exist integers with product $ac$ and sum $b$.
Caveat: Almost every quadratic polynomial you can write down is irreducible with respect to the rational numbers, meaning it cannot be split into two linear factors with rational coefficients. For instance, $x^2 + 2x + 3$ is irreducible with respect to the rational numbers since there are no integers with product $1 cdot 3 = 3$ and sum $2$.
answered Aug 2 at 17:50
N. F. Taussig
38k93053
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider what happens when you multiply two linear polynomials.
beginalign*
(4x - 5)(3x + 8) & = 4x(3x + 8) - 5(3x + 8) && textapply the distributive law\
& = colorblue12x^2 + colorgreen32x - colorgreen15x - colorblue40 && textapply the distributive law\
& = colorblue12x^2 + colorgreen17x - colorblue40 && textcombine like terms
endalign*
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients that combine to form the linear coefficient of the product, that is,
$$(colorblue12)(colorblue-40) = (colorgreen32)( colorgreen-15)$$
To factor $12x^2 + 17x - 40$, we wish to carry out these steps in reverse.
beginalign*
12x^2 + colorgreen17x - 40 & = 12x^2 + 32x - 15x - 40 && textsplit the linear term\
& = 4x(3x + 8) - 5(3x + 8) && textfactor by grouping\
& = (4x - 5)(3x + 8) && textextract the common factor
endalign*
The observation we made above enables us to split the linear term. To do so, we must find two numbers with product $(colorblue12)(colorblue-40) = -480$ that have sum $colorgreen17$. We can do so by listing the factor pairs of $-480$, then finding the pair in which the sum of the factors is $colorgreen17$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
(1)(-480) & -479 & (-1)(480) & 479\
(2)(-240) & -238 & (-2)(240) & 238\
(3)(-160) & -157 & (-3)(160) & 157\
(4)(-120) & -116 & (-4)(120) & 116\
(5)(-96) & -91 & (-5)(96) & 91\
(6)(-80) & -74 & (-6)(80) & 74\
(8)(-60) & -52 & (-8)(60) & 52\
(10)(-48) & -38 & (-10)(48) & 38\
(12)(-40) & -28 & (-12)(40) & 28\
(15)(-32) & -17 & (colorgreen-15)(colorgreen32) & colorgreen17\
(20)(-24) & -4 & (-20)(24) & 4
endarray
The desired factors are $colorgreen32$ and $colorgreen-15$. Once we split the linear term, we extract the common factor from the first two terms and the common factor from the last two terms, which is called factoring by grouping. Finally, we extract the common linear factor to complete the factorization.
Let's apply this technique to your example.
$$5c^2 - 52c + 20$$
To split the linear term, we must find two numbers with product $colorblue5 cdot colorblue20 = 100$ and sum $colorgreen-52$. We list the factor pairs of $100$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
1 cdot 100 & 101 & (-1)(-100) & -101\
2 cdot 50 & 52 & (colorgreen-2)(colorgreen-50) & colorgreen-52\
4 cdot 25 & 29 & (-4)(-25) & -29\
5 cdot 20 & 25 & (-5)(-20) & -25\
10 cdot 10 & 20 & (-10)(-10) & -20
endarray
The desired factors are $colorgreen-2$ and $colorgreen-50$. Hence,
beginalign*
5c^2 - 52c + 20 & = 5c^2 - 2c - 50c + 20 && textsplit the linear term\
& = c(5c - 2) - 10(5c - 2) && textfactor by grouping\
& = (c - 10)(5c - 2) && textextract the common factor
endalign*
More generally, a factorization
$$ax^2 + bx + c = (rx + s)(tx + u)$$
of a quadratic polynomial $ax^2 + bx + c$ with respect to the rational numbers can be found if $a$, $b$, and $c$ are integers such that there exist integers with product $ac$ and sum $b$.
Caveat: Almost every quadratic polynomial you can write down is irreducible with respect to the rational numbers, meaning it cannot be split into two linear factors with rational coefficients. For instance, $x^2 + 2x + 3$ is irreducible with respect to the rational numbers since there are no integers with product $1 cdot 3 = 3$ and sum $2$.
answered Aug 2 at 17:50
N. F. Taussig
38k93053
Consider what happens when you multiply two linear polynomials.
beginalign*
(4x - 5)(3x + 8) & = 4x(3x + 8) - 5(3x + 8) && textapply the distributive law\
& = colorblue12x^2 + colorgreen32x - colorgreen15x - colorblue40 && textapply the distributive law\
& = colorblue12x^2 + colorgreen17x - colorblue40 && textcombine like terms
endalign*
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients that combine to form the linear coefficient of the product, that is,
$$(colorblue12)(colorblue-40) = (colorgreen32)( colorgreen-15)$$
To factor $12x^2 + 17x - 40$, we wish to carry out these steps in reverse.
beginalign*
12x^2 + colorgreen17x - 40 & = 12x^2 + 32x - 15x - 40 && textsplit the linear term\
& = 4x(3x + 8) - 5(3x + 8) && textfactor by grouping\
& = (4x - 5)(3x + 8) && textextract the common factor
endalign*
The observation we made above enables us to split the linear term. To do so, we must find two numbers with product $(colorblue12)(colorblue-40) = -480$ that have sum $colorgreen17$. We can do so by listing the factor pairs of $-480$, then finding the pair in which the sum of the factors is $colorgreen17$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
(1)(-480) & -479 & (-1)(480) & 479\
(2)(-240) & -238 & (-2)(240) & 238\
(3)(-160) & -157 & (-3)(160) & 157\
(4)(-120) & -116 & (-4)(120) & 116\
(5)(-96) & -91 & (-5)(96) & 91\
(6)(-80) & -74 & (-6)(80) & 74\
(8)(-60) & -52 & (-8)(60) & 52\
(10)(-48) & -38 & (-10)(48) & 38\
(12)(-40) & -28 & (-12)(40) & 28\
(15)(-32) & -17 & (colorgreen-15)(colorgreen32) & colorgreen17\
(20)(-24) & -4 & (-20)(24) & 4
endarray
The desired factors are $colorgreen32$ and $colorgreen-15$. Once we split the linear term, we extract the common factor from the first two terms and the common factor from the last two terms, which is called factoring by grouping. Finally, we extract the common linear factor to complete the factorization.
Let's apply this technique to your example.
$$5c^2 - 52c + 20$$
To split the linear term, we must find two numbers with product $colorblue5 cdot colorblue20 = 100$ and sum $colorgreen-52$. We list the factor pairs of $100$.
beginarrayc c c c
textfactorization & textfactor sum & textfactorization & textfactor sum\ hline
1 cdot 100 & 101 & (-1)(-100) & -101\
2 cdot 50 & 52 & (colorgreen-2)(colorgreen-50) & colorgreen-52\
4 cdot 25 & 29 & (-4)(-25) & -29\
5 cdot 20 & 25 & (-5)(-20) & -25\
10 cdot 10 & 20 & (-10)(-10) & -20
endarray
The desired factors are $colorgreen-2$ and $colorgreen-50$. Hence,
beginalign*
5c^2 - 52c + 20 & = 5c^2 - 2c - 50c + 20 && textsplit the linear term\
& = c(5c - 2) - 10(5c - 2) && textfactor by grouping\
& = (c - 10)(5c - 2) && textextract the common factor
endalign*
More generally, a factorization
$$ax^2 + bx + c = (rx + s)(tx + u)$$
of a quadratic polynomial $ax^2 + bx + c$ with respect to the rational numbers can be found if $a$, $b$, and $c$ are integers such that there exist integers with product $ac$ and sum $b$.
Caveat: Almost every quadratic polynomial you can write down is irreducible with respect to the rational numbers, meaning it cannot be split into two linear factors with rational coefficients. For instance, $x^2 + 2x + 3$ is irreducible with respect to the rational numbers since there are no integers with product $1 cdot 3 = 3$ and sum $2$.
answered Aug 2 at 17:50
N. F. Taussig
38k93053
answered Aug 2 at 17:50
N. F. Taussig
38k93053
answered Aug 2 at 17:50
N. F. Taussig
38k93053
answered Aug 2 at 17:50
N. F. Taussig
38k93053
38k93053
add a comment |Â
add a comment |Â
up vote
0
down vote
There's always completing the square and/or the quadratic formula.
And bear in mind not all quadratic expression factor to rational components.
But:
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
Let's explain:
So if we solve $5c^2 - 52c + 20 = 0$ by completing the square (or quadratic formula) we get:
$5c^2 - 52c + 20 = 0$
$c^2 - frac 525c + 4 = 0$
$c^2 - frac 525c = -4$
$c^2 - frac 525c + (frac 265)^2 $
$(c - frac 265)^2 = = (frac 265)^2 - 4=frac 4*13^225 - frac 4*2525= frac 4(13^2 - 5^2)25 = frac 4*14425=(frac 2*125)^2$
$c -frac 265 = pm frac 245$
$c = frac 265 pm 245$
$c = frac 25$ or $c = 10$ so
.....
Now
Suppose $5c^2 - 52c + 20$ factored into $(mc + u)(nc + v)=mn(c +frac um)(c+frac vn)$. Then $5c^2 - 52c + 20 = 0 iff (c +frac um)(c+frac vn) = 0 iff c= -frac um$ or $c = -frac vn$.
So we know that the solutions are $c = frac 25$ and $c = 10$ so we must have that $5c^2 - 52c + 20$ factors to $5(c-frac 25)(c-10)$ or $(5c - 2)(c-20)$.
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
....
That's how you do it if you just can't see anything else.
In general you can make educated guesses:
If $5c^2 + 52c + 20 = (mc + u)(nc+ v)$ then
$mn = 5$
$un + vm = -52$
$uv = 20$.
If this behaves "nicely" (which you have no reason to assume it will!) then $mn = 5$ means $m,n = 5,1$.
And $uv = 20$ so $u,v =pm 1,20, pm 2,10, pm 4,5$
We have $u + 5v = -52$. The mean $u = -2, - 7 ,-12....$ and $v = -10, -9, -8....$ etc. The only choses we have that work are $-2,-10$.
So it was a very educated guess that $(5c -2)(c-10) =5c^2 + (-2-5*10)c +(-2*-10) = 5c^2 -52c + 20$.
answered Aug 2 at 18:52
fleablood
60.1k22575
add a comment |Â
up vote
0
down vote
There's always completing the square and/or the quadratic formula.
And bear in mind not all quadratic expression factor to rational components.
But:
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
Let's explain:
So if we solve $5c^2 - 52c + 20 = 0$ by completing the square (or quadratic formula) we get:
$5c^2 - 52c + 20 = 0$
$c^2 - frac 525c + 4 = 0$
$c^2 - frac 525c = -4$
$c^2 - frac 525c + (frac 265)^2 $
$(c - frac 265)^2 = = (frac 265)^2 - 4=frac 4*13^225 - frac 4*2525= frac 4(13^2 - 5^2)25 = frac 4*14425=(frac 2*125)^2$
$c -frac 265 = pm frac 245$
$c = frac 265 pm 245$
$c = frac 25$ or $c = 10$ so
.....
Now
Suppose $5c^2 - 52c + 20$ factored into $(mc + u)(nc + v)=mn(c +frac um)(c+frac vn)$. Then $5c^2 - 52c + 20 = 0 iff (c +frac um)(c+frac vn) = 0 iff c= -frac um$ or $c = -frac vn$.
So we know that the solutions are $c = frac 25$ and $c = 10$ so we must have that $5c^2 - 52c + 20$ factors to $5(c-frac 25)(c-10)$ or $(5c - 2)(c-20)$.
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
....
That's how you do it if you just can't see anything else.
In general you can make educated guesses:
If $5c^2 + 52c + 20 = (mc + u)(nc+ v)$ then
$mn = 5$
$un + vm = -52$
$uv = 20$.
If this behaves "nicely" (which you have no reason to assume it will!) then $mn = 5$ means $m,n = 5,1$.
And $uv = 20$ so $u,v =pm 1,20, pm 2,10, pm 4,5$
We have $u + 5v = -52$. The mean $u = -2, - 7 ,-12....$ and $v = -10, -9, -8....$ etc. The only choses we have that work are $-2,-10$.
So it was a very educated guess that $(5c -2)(c-10) =5c^2 + (-2-5*10)c +(-2*-10) = 5c^2 -52c + 20$.
answered Aug 2 at 18:52
fleablood
60.1k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There's always completing the square and/or the quadratic formula.
And bear in mind not all quadratic expression factor to rational components.
But:
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
Let's explain:
So if we solve $5c^2 - 52c + 20 = 0$ by completing the square (or quadratic formula) we get:
$5c^2 - 52c + 20 = 0$
$c^2 - frac 525c + 4 = 0$
$c^2 - frac 525c = -4$
$c^2 - frac 525c + (frac 265)^2 $
$(c - frac 265)^2 = = (frac 265)^2 - 4=frac 4*13^225 - frac 4*2525= frac 4(13^2 - 5^2)25 = frac 4*14425=(frac 2*125)^2$
$c -frac 265 = pm frac 245$
$c = frac 265 pm 245$
$c = frac 25$ or $c = 10$ so
.....
Now
Suppose $5c^2 - 52c + 20$ factored into $(mc + u)(nc + v)=mn(c +frac um)(c+frac vn)$. Then $5c^2 - 52c + 20 = 0 iff (c +frac um)(c+frac vn) = 0 iff c= -frac um$ or $c = -frac vn$.
So we know that the solutions are $c = frac 25$ and $c = 10$ so we must have that $5c^2 - 52c + 20$ factors to $5(c-frac 25)(c-10)$ or $(5c - 2)(c-20)$.
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
....
That's how you do it if you just can't see anything else.
In general you can make educated guesses:
If $5c^2 + 52c + 20 = (mc + u)(nc+ v)$ then
$mn = 5$
$un + vm = -52$
$uv = 20$.
If this behaves "nicely" (which you have no reason to assume it will!) then $mn = 5$ means $m,n = 5,1$.
And $uv = 20$ so $u,v =pm 1,20, pm 2,10, pm 4,5$
We have $u + 5v = -52$. The mean $u = -2, - 7 ,-12....$ and $v = -10, -9, -8....$ etc. The only choses we have that work are $-2,-10$.
So it was a very educated guess that $(5c -2)(c-10) =5c^2 + (-2-5*10)c +(-2*-10) = 5c^2 -52c + 20$.
answered Aug 2 at 18:52
fleablood
60.1k22575
There's always completing the square and/or the quadratic formula.
And bear in mind not all quadratic expression factor to rational components.
But:
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
Let's explain:
So if we solve $5c^2 - 52c + 20 = 0$ by completing the square (or quadratic formula) we get:
$5c^2 - 52c + 20 = 0$
$c^2 - frac 525c + 4 = 0$
$c^2 - frac 525c = -4$
$c^2 - frac 525c + (frac 265)^2 $
$(c - frac 265)^2 = = (frac 265)^2 - 4=frac 4*13^225 - frac 4*2525= frac 4(13^2 - 5^2)25 = frac 4*14425=(frac 2*125)^2$
$c -frac 265 = pm frac 245$
$c = frac 265 pm 245$
$c = frac 25$ or $c = 10$ so
.....
Now
Suppose $5c^2 - 52c + 20$ factored into $(mc + u)(nc + v)=mn(c +frac um)(c+frac vn)$. Then $5c^2 - 52c + 20 = 0 iff (c +frac um)(c+frac vn) = 0 iff c= -frac um$ or $c = -frac vn$.
So we know that the solutions are $c = frac 25$ and $c = 10$ so we must have that $5c^2 - 52c + 20$ factors to $5(c-frac 25)(c-10)$ or $(5c - 2)(c-20)$.
In general:
if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = frac -b
+sqrtb^2 -4ac2a$ or $x = k_2 = frac -b - sqrtb^2 - 4ac2a$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.
....
That's how you do it if you just can't see anything else.
In general you can make educated guesses:
If $5c^2 + 52c + 20 = (mc + u)(nc+ v)$ then
$mn = 5$
$un + vm = -52$
$uv = 20$.
If this behaves "nicely" (which you have no reason to assume it will!) then $mn = 5$ means $m,n = 5,1$.
And $uv = 20$ so $u,v =pm 1,20, pm 2,10, pm 4,5$
We have $u + 5v = -52$. The mean $u = -2, - 7 ,-12....$ and $v = -10, -9, -8....$ etc. The only choses we have that work are $-2,-10$.
So it was a very educated guess that $(5c -2)(c-10) =5c^2 + (-2-5*10)c +(-2*-10) = 5c^2 -52c + 20$.
answered Aug 2 at 18:52
fleablood
60.1k22575
answered Aug 2 at 18:52
fleablood
60.1k22575
answered Aug 2 at 18:52
fleablood
60.1k22575
answered Aug 2 at 18:52
fleablood
60.1k22575
60.1k22575
add a comment |Â
add a comment |Â
up vote
0
down vote
$$beginalign
5c^2 - 52c + 20 &= frac55cdotbig(5c^2 - 52c + 20big)\
&= frac25c^2 - 260c + 1005\
&= frac(5c)^2 -52(5c) + 1005\
&= fract^2 -52t + 1005text where t=5c\
endalign$$
Can you take it from there?
answered 2 days ago
John Joy
5,86511526
add a comment |Â
up vote
0
down vote
$$beginalign
5c^2 - 52c + 20 &= frac55cdotbig(5c^2 - 52c + 20big)\
&= frac25c^2 - 260c + 1005\
&= frac(5c)^2 -52(5c) + 1005\
&= fract^2 -52t + 1005text where t=5c\
endalign$$
Can you take it from there?
answered 2 days ago
John Joy
5,86511526
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$beginalign
5c^2 - 52c + 20 &= frac55cdotbig(5c^2 - 52c + 20big)\
&= frac25c^2 - 260c + 1005\
&= frac(5c)^2 -52(5c) + 1005\
&= fract^2 -52t + 1005text where t=5c\
endalign$$
Can you take it from there?
answered 2 days ago
John Joy
5,86511526
$$beginalign
5c^2 - 52c + 20 &= frac55cdotbig(5c^2 - 52c + 20big)\
&= frac25c^2 - 260c + 1005\
&= frac(5c)^2 -52(5c) + 1005\
&= fract^2 -52t + 1005text where t=5c\
endalign$$
Can you take it from there?
answered 2 days ago
John Joy
5,86511526
answered 2 days ago
John Joy
5,86511526
answered 2 days ago
John Joy
5,86511526
answered 2 days ago
John Joy
5,86511526
5,86511526
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