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)(x1) 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. 
January 3rd, 2019, 08:15 AM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Quote:
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:
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(u14)(u36) &= 0 &(\text{not easy to spot because of the large numbers}) \\ \tfrac19(9w14)(9w36) &= 0 \\ (9w14)\frac{9w36}{9} &= 0 \\ (9w14)(w4) &= 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 squareroot 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 timeywimey stuff.  Quote:
Dan  
January 3rd, 2019, 11:21 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677  
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  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:
 
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]) 

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coefficients, equations, factoring, leading, quadratic 
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