How can I factorize this quadratic expression

Question

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?

Answer 1

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: $$4x^2 - 4x - 15 = 4x^2 + 6x - 10x - 15 = 2x(2x+3) - 5(2x+3) = (2x-5)(2x+3).$$

Answer 2

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.

Answer 3

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

Answer 4

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.