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Going by the exercises of a book I have been factorizing quadratic equations the following way, let's say I have:

$$ {x^2 - 7x + 12 = 0} $$

I know that

$$ {a \times b = 12 \\ \text{ and } \\ a + b = -7} $$

So it's easy enough to figure out that -3 and -4 will do.

The result is obviously:

$$ {(x - 3)(x - 4) = 0} $$

I a trying to apply the same logic to the following:

$$ {4x^2 - 4x - 15 = 0} $$

so

$$ {a \times b = -15 \\ \text{ and } \\ a + b = -4} $$

There is no obvious answer that I can think of, I guess I could solve the equation system and get an a and a and a b but this makes no sense in this context since it would basically only add complexity to the original context.

Am I missing something obvious here?

How can I factorize this? Should modify the original equation some way so that it's possible?

Trufa
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  • hint : let $(2x-a)(2x-b)=4x^2-4x-15$ now find a,b – Khosrotash Apr 13 '15 at 04:37
  • @daryakhosrotash ok, I'm trying... thanks – Trufa Apr 13 '15 at 04:39
  • $$(2x-a)(2x-b)=4x^2-2x(a+b)+ab\so\-2(a+b)=-4\ab=-15$$ – Khosrotash Apr 13 '15 at 04:40
  • @daryakhosrotash ohh, I think this is making sense now.... – Trufa Apr 13 '15 at 04:44
  • @daryakhosrotash: that works this time, but it could be $(4x-a)(x-b)$ that works the next time. You can change that to $(2x-\frac a2)(2x-2b)$, but if $a$ is odd the division doesn't come out even. – Ross Millikan Apr 13 '15 at 04:44
  • it can be transformed ,whenever you want .see this $$(2x-3)(2x-5)=\2(2x-3)(x-\frac{5}{2})=\(4x-6)(x-\frac{5}{2})$$ – Khosrotash Apr 13 '15 at 04:47
  • We have that if $ax^2+bx+c$ is the polynomial you want to factorize (and $a\neq 0$), then $x_1+x_2=-\frac{b}{a}, x_1x_2=\frac{c}{a}$ and $ax^2+bx+c=a(x-x_1)(x-x_2)$. You have done a whole lot of mistakes (almost everywhere. Through wrong reasoning you got the correct factorization) in your attempts (sadly no one has mentioned it here yet), so see the facts I've written here and closely find your mistakes. – user26486 Apr 13 '15 at 15:01

5 Answers5

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In the case of a non-one coefficient in front of the $x^2$ term, like $ax^2 + bx + c$, you have to find two numbers that add to $b$ and that multiply to $ac$ instead of to $c$. So for $4x^2 - 4x - 15$ we need two numbers that multiply to $4\times(-15) = -60$ and add to $-4$; $6$ and $-10$ work. Then: $$4x^2 - 4x - 15 = 4x^2 + 6x - 10x - 15 = 2x(2x+3) - 5(2x+3) = (2x-5)(2x+3).$$

Unit
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  • Ok, didn't know that, but 60 has a lot of factors and it would take (at least) me a while to find them, they wouldn't come intuitively. – Trufa Apr 13 '15 at 04:46
  • Since $4$ is small compared to $60$, your pair of factors is not very far apart. You could start looking around the "middle": $\sqrt{60}$, which is between $7$ and $8$ (because $60$ is between $49$ and $64$). Going down from $8$ we test for divisors of $60$ until we get to $6$, which works. – Unit Apr 13 '15 at 18:15
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Your rule for finding $a$ and $b$ depends on the fact that the coefficient of the $x^2$ term is $1$. In that case you are expanding $(x-a)(x-b)=x^2-(a+b)x+ab$. If the leading coefficient is not $1$, you can divide it out. In your example, that gives $4(x^2-x-\frac {15}4)$ and now you can use your technique.

Ross Millikan
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We can always use the quadratic formula to find the roots of such an equation. Recall that $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Hence, \begin{align*} x&=\frac{-(-4)\pm\sqrt{4^2-4\cdot4\cdot(-15)}}{2\cdot 4}\\ &=\frac{4\pm\sqrt{16+16\cdot 15}}{8}\\ &=\frac{4\pm16}{8}\\ &=\frac{1}{2}\pm2. \end{align*} Therefore, the roots of the equation are $$x=\frac{5}{2} \ \text{or} \ x=\frac{-3}{2}.$$ From this, we can see that $$x-\frac{5}{2}=0 \ \Longrightarrow \ 2x-5=0$$ and $$x+\frac{3}{2}=0 \ \Longrightarrow \ 2x+3=0.$$ This means that we can rewrite the quadratic equation in the form $$4x^2-4x-15=(2x-5)(2x+3).$$

  • Thanks, I am aware of this, but this section of the book in particular tries to teach factorization as a method for solving. Thank you very much anyway! – Trufa Apr 13 '15 at 04:56
  • You're welcome. As a high school student I was always really bad at factorizing quadratics by sort of guessing at the factors. I always used the quadratic equation in the end. –  Apr 13 '15 at 05:03
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Hint $\,\ 4x^2\!-4x-15\, =\, (2x)^2\!-2(2x)-15\, = \color{#c00}{X^2}-2X-15,\ $ for $\, X = 2x$

Remark $\ $ The idea generalizes to a way to reduce polynomial factorization to the case where the leading coefficient $\color{#c00}{=\bf 1}$ (i.e. is $\rm\color{#c00}{monic})$, see the AC method. It is very handy for such problems.

Community
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Bill Dubuque
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Use what is known as the "$ac$ method". Multiply $a$ by $c$ to get $-60 (4*-15)$

Thus you find two numbers $(p,q)$ whose product is $-60$ and whose sum is $-4$. In this case, a very quick check of the factors of $60$ will show that $p=6, q=-10$.

Now, if $(ax+b)(cx+d)=4x^2-4x-15, ab=6, cd=-10, ac=4, bd=-15$, based on what we just worked out. From this, we can deduce $a=2, b=3,c=2,d=-5$, thus we get $(2x+3)(2x-5)$ and solve for $x=-\frac{3}{2}, x=\frac{5}{2}$

Rhys Hughes
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