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

 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 3rd, 2019, 04:47 AM #1 Newbie   Joined: Jan 2019 From: UK Posts: 1 Thanks: 0 I would like some help with factoring quadratic equations with leading coefficients. 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. January 3rd, 2019, 07:17 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2337 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. Thanks from topsquark January 3rd, 2019, 08:15 AM   #3
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 552

Quote:
 Originally Posted by SonicRainboom 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. January 3rd, 2019, 09:30 AM   #4
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,700
Thanks: 2682

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by SonicRainboom 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 by v8archie; January 3rd, 2019 at 09:33 AM. January 3rd, 2019, 10:09 AM   #5
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,345
Thanks: 986

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by JeffM1 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 January 3rd, 2019, 11:21 AM   #6
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,101
Thanks: 1677

Quote:
 Originally Posted by topsquark ... (I'm rabidly against that policy, by the way.) -Dan
RRRR ... MAD DOG!   January 3rd, 2019, 11:27 AM   #7
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,700
Thanks: 2682

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by topsquark 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. January 3rd, 2019, 11:33 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 That's gotta be Texas Toothbrush! January 3rd, 2019, 07:46 PM   #9
Newbie

Joined: Jan 2019
From: Dayton

Posts: 2
Thanks: 0

Math Focus: algebra, calculus, numerical methods Quote:
 Originally Posted by SonicRainboom 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. January 11th, 2019, 07:24 AM #10 Newbie   Joined: Jan 2019 From: Arrakis Posts: 7 Thanks: 0 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]) Tags coefficients, equations, factoring, leading, quadratic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Popcorn23 Algebra 1 December 13th, 2015 02:41 AM mms Algebra 23 February 14th, 2014 04:41 AM watkd Algebra 18 September 6th, 2010 02:34 PM sivela Algebra 4 January 27th, 2010 01:42 PM JamesDorman Algebra 5 December 14th, 2009 03:51 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      