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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?

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  • $\begingroup$ hint : let $(2x-a)(2x-b)=4x^2-4x-15$ now find a,b $\endgroup$ – Khosrotash Apr 13 '15 at 4:37
  • $\begingroup$ @daryakhosrotash ok, I'm trying... thanks $\endgroup$ – Trufa Apr 13 '15 at 4:39
  • $\begingroup$ $$(2x-a)(2x-b)=4x^2-2x(a+b)+ab\\so\\-2(a+b)=-4\\ab=-15$$ $\endgroup$ – Khosrotash Apr 13 '15 at 4:40
  • $\begingroup$ @daryakhosrotash ohh, I think this is making sense now.... $\endgroup$ – Trufa Apr 13 '15 at 4:44
  • $\begingroup$ @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. $\endgroup$ – Ross Millikan Apr 13 '15 at 4:44
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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: $$4x^2 - 4x - 15 = 4x^2 + 6x - 10x - 15 = 2x(2x+3) - 5(2x+3) = (2x-5)(2x+3).$$

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  • $\begingroup$ 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. $\endgroup$ – Trufa Apr 13 '15 at 4:46
  • $\begingroup$ 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. $\endgroup$ – 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.

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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 $$4x^2-4x-15=(2x-5)(2x+3).$$

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  • $\begingroup$ 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! $\endgroup$ – Trufa Apr 13 '15 at 4:56
  • $\begingroup$ 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. $\endgroup$ – user230944 Apr 13 '15 at 5:03
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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.

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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 $x=-\frac{3}{2}, x=\frac{5}{2}$

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