I would like some help with factoring quadratic equations with leading coefficients.

Jan 2019
1
0
UK
Hello. I seem to have some problems factoring quadratic equations when the leading term has a coefficient.

Here is an example

3w -7 = sqr(8w -7)
(3w -7)^2 = (sqr(8w -7))^2
9w^2 -42w +49 = 8w -7
9w^2 -50w +56 = 0

Ok so here we have a standard form quadratic equation with a leading coefficient. I know that without a leading coefficient I'm looking for two numbers that would add together to make -50 and multiply together to make 56, I'm comfortable with that.

But what do I do when there is a leading coefficient that can not be factored out such as the 9 above?

I was told that with a coefficient I should find two numbers that add up to make the middle term and multiply together to make the last term * the leading coefficient (56 * 9 = 504).

In this case it would be -14 and -36

Here is my confusion... because that doesn't work! The answer is

(9w -14)(w -4) = 0

How does one get to -14 and 4?

I can see it will works since -4 * 9 = -36 and then -36 * -14 = 50 but starting from standard form what method do I use to find -14 and -4?

Here is another example

2x^3 -7x^2 +5x
= x(2x -5)(x-1)

I can see that
x(2x^2 -7x +5)

But then how do you get -5 and -1?

Once again I thought I was looking for a numbers that add up to make -7 and multiplies to make 2* 5 = 10 so I would have got -5 and -2... but the answer is -5 and -1.

Could somebody help me understand the method from which I find those two numbers from standard form? Thank you.
 

skipjack

Forum Staff
Dec 2006
21,473
2,466
You were given correct advice.

If, given ax² + bx + c, you find m and n such that m + n = b and mn = ac,

(ax + m)(ax + n)/a = (a²x² + a(m + n)x + mn)/a = ax² + bx + c.

As mn = ac, (ax + m)(ax + n)/a can be simplified to give an answer without fractions.
 
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May 2016
1,310
551
USA
Hello. I seem to have some problems factoring quadratic equations when the leading term has a coefficient.

Here is an example

3w -7 = sqr(8w -7)
(3w -7)^2 = (sqr(8w -7))^2
9w^2 -42w +49 = 8w -7
9w^2 -50w +56 = 0

Ok so here we have a standard form quadratic equation with a leading coefficient. I know that without a leading coefficient I'm looking for two numbers that would add together to make -50 and multiply together to make 56, I'm comfortable with that.

But what do I do when there is a leading coefficient that can not be factored out such as the 9 above?

I was told that with a coefficient I should find two numbers that add up to make the middle term and multiply together to make the last term * the leading coefficient (56 * 9 = 504).

In this case it would be -14 and -36

Here is my confusion... because that doesn't work! The answer is

(9w -14)(w -4) = 0

How does one get to -14 and 4?

I can see it will works since -4 * 9 = -36 and then -36 * -14 = 50 but starting from standard form what method do I use to find -14 and -4?

Here is another example

2x^3 -7x^2 +5x
= x(2x -5)(x-1)

I can see that
x(2x^2 -7x +5)

But then how do you get -5 and -1?

Once again I thought I was looking for a numbers that add up to make -7 and multiplies to make 2* 5 = 10 so I would have got -5 and -2... but the answer is -5 and -1.

Could somebody help me understand the method from which I find those two numbers from standard form? Thank you.
There are several methods for solving a quadratic equation. The factoring method is sometimes the fastest, but more usually is the slowest.

Have you been taught the other ways to solve a quadratic equation? What I say to my students is: try factoring first, but if that does not work quickly, switch to the quadratic formula.
 
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v8archie

Math Team
Dec 2013
7,710
2,679
Colombia
Hello. I seem to have some problems factoring quadratic equations when the leading term has a coefficient.
For this case (with large numbers) I'd probably go with the quadratic formula, but the best manual method I know is as follows:

Given $$f(x) = ax^2 + bx +c$$
we multiply the entire expression by $\frac{a}{a}$. This turns the first term into an exact square. \begin{align}f(x) &= \tfrac{a}{a}(ax^2 + bx + c) \\ &= \tfrac1a(a^2x^2 + abx + ac) \\ &= \tfrac1a\big((ax)^2 + b(ax) + ac\big)\end{align}
The expression is now a quadratic in $(ax)$, but with a leading coefficient of $1$. To make this clearer we can write $u=ax$ and then the equation becomes $$f(u) = \tfrac1a(u^2 + bu + ac)$$
We can now factor this in our favourite way to get $$f(u) = \tfrac1a(u+p)(u+q)$$
Now we replace the $u$ with $ax$ and get $$f(x) = \tfrac1a(ax+p)(ax+q)$$
If there are any rational roots (which there presumably are since we managed to factor the equation without using the formula), at least one of $p$ and $q$ is divisible by $a$, so we can simplify the above.

For example:
\begin{align}
9w^2 - 50w +56 &= 0 \\
\tfrac99(9w^2 - 50w + 56) &= 0 \\
\tfrac19(9^2w^2 - 9(50w) + 504) &= 0 \\
\tfrac19\big((9w)^2 - 50(9w) + 504\big) &= 0 \\
u=9w \implies \tfrac19(u^2 - 50u + 504) &= 0 \\
\tfrac19(u-14)(u-36) &= 0 &(\text{not easy to spot because of the large numbers}) \\
\tfrac19(9w-14)(9w-36) &= 0 \\
(9w-14)\frac{9w-36}{9} &= 0 \\
(9w-14)(w-4) &= 0
\end{align}

In the case that the leading coefficient is a square number (such as in this example) and the second coefficient is divisible by the square-root of that coefficient (not as in this case), you can avoid multiplying by the leading coefficient, but it isn't necessary to do so.
 
Last edited:

topsquark

Math Team
May 2013
2,510
1,049
The Astral plane
Have you been taught the other ways to solve a quadratic equation? What I say to my students is: try factoring first, but if that does not work quickly, switch to the quadratic formula.
There are a large number of instructors that would say "forget about the quadratic formula... you've all got graphing calculators." (I'm rabidly against that policy, by the way.)

-Dan
 

v8archie

Math Team
Dec 2013
7,710
2,679
Colombia
There are a large number of instructors that would say "forget about the quadratic formula... you've all got graphing calculators." (I'm rabidly against that policy, by the way.)

-Dan
Symbolic calculators, fine - but graphing? You can't find exact answers on a graph.
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
That's gotta be Texas Toothbrush!
 
Jan 2019
2
0
Dayton
Hello. I seem to have some problems factoring quadratic equations when the leading term has a coefficient.

Here is an example

3w -7 = sqr(8w -7)
(3w -7)^2 = (sqr(8w -7))^2
9w^2 -42w +49 = 8w -7
9w^2 -50w +56 = 0

Ok so here we have a standard form quadratic equation with a leading coefficient. I know that without a leading coefficient I'm looking for two numbers that would add together to make -50 and multiply together to make 56, I'm comfortable with that.

But what do I do when there is a leading coefficient that can not be factored out such as the 9 above?

I was told that with a coefficient I should find two numbers that add up to make the middle term and multiply together to make the last term * the leading coefficient (56 * 9 = 504).

In this case it would be -14 and -36

Here is my confusion... because that doesn't work! The answer is

(9w -14)(w -4) = 0

How does one get to -14 and 4?

I can see it will works since -4 * 9 = -36 and then -36 * -14 = 50 but starting from standard form what method do I use to find -14 and -4?

Here is another example

2x^3 -7x^2 +5x
= x(2x -5)(x-1)

I can see that
x(2x^2 -7x +5)

But then how do you get -5 and -1?

Once again I thought I was looking for a numbers that add up to make -7 and multiplies to make 2* 5 = 10 so I would have got -5 and -2... but the answer is -5 and -1.

Could somebody help me understand the method from which I find those two numbers from standard form? Thank you.
You will need to split the middle term -50W into two terms in such a way that it lends us to get a common expression out from the new 4 terms quadratic expression. Check out this link for steps on how the middle term is broken.
 
Jan 2019
7
0
Arrakis
If I were you in this situation, I would definitely switch to the quadratic formula as a method to factor these quadratic equations.

Another method, however, assuming your polynomial is in the form of ax^k+bx+c, to factor the polynomial, you would:

1. Multiply the coefficients a and c. (If your polynomial is 8x^2+10x+3, that would make 24.)

2. Find a factor of ac that add together to make b. If you can find this, then the polynomial is factorable. (24=6x4; 6+4=10)

3. Break apart b into its components you just found. (8x^2+6x+4x+3)

4. Factor by grouping. (2x[4x+3]+[4x+3]; [2x+1][4x+3])