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

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 January 3rd, 2019, 06:17 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,975 Thanks: 2226 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, 07:15 AM   #3
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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, 08:30 AM   #4
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 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 08:33 AM.

January 3rd, 2019, 09:09 AM   #5
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 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, 10:21 AM   #6
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 Originally Posted by topsquark ... (I'm rabidly against that policy, by the way.) -Dan

January 3rd, 2019, 10:27 AM   #7
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 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, 10:33 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 That's gotta be Texas Toothbrush!
January 3rd, 2019, 06:46 PM   #9
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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, 06:24 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])

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