2
$\begingroup$

Expanding very simple process, just open round brackets and multiply each with each for example

$(x-6)(x+3) = x^2 + 3x - 6x -18 = x^2 -3x -18$

Factoring is confused me a little for example I have $x^2 -3x -18$ I try to brute force all combinations in my mind and get the result like this $(x-6)(x+3)$ Cold you please recommend some algorithm, how to find factoring in the easy way.

Thanks.

$\endgroup$
1
$\begingroup$

You can do this:

$$x^2-3x-18=\left(x-\frac 32 x\right)^2-\frac 94-18$$ $$=\left(x-\frac 32 \right)^2-\frac {81}4$$ $$=\left(x-\frac 32 -\sqrt{\frac{81}4}\right)\left(x-\frac 32 x+\sqrt{\frac{81}4}\right)$$ $$=\left(x-\frac 32 -\frac{9}2\right)\left(x-\frac 32 +\frac{9}2\right)$$ $$=(x-6)(x+3).$$

$\endgroup$
1
$\begingroup$

Here's an algorithm to factor any quadratic (so $a\neq 0$):

\begin{align}ax^2+bx+c&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right)\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}\right) &(\Delta=b^2-4ac)\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\left(\frac{\sqrt{\Delta}}{2a}\right)^2\right)\\&=a\left(x+\frac{b-\sqrt{\Delta}}{2a}\right)\left(x+\frac{b+\sqrt{\Delta}}{2a}\right)\end{align}

Note that we used $\sqrt{\Delta}$. If you know complex numbers, that shouldn't be a problem. If not, note that when $\Delta<0$, the equation $ax^2+bx+c=0$ has no solutions in $\mathbb{R}$ because it's equivalent to $$\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}=0$$ which has clearly no solutions in $\mathbb{R}$ as $\left(x+\frac{b}{2a}\right)^2\ge 0$ and $-\frac{\Delta}{4a^2}>0$. Thus, if $\Delta<0$, you can't factor over $\mathbb{R}$.

$\endgroup$
1
$\begingroup$

Here is another alternative

$$ \begin{align} x^2-3x-18 &= \frac{1}{4}\bigg(4x^2-4\cdot 3x-4\cdot 18\bigg )\\ &= \frac{1}{4}\bigg((2x)^2-2\cdot 3(2x)-72\bigg )\\ \end{align}$$

Now suppose we define $t=2x$ $$\begin{align} \frac{1}{4}\bigg(t^2-2\cdot 3t-72\bigg )&= \frac{1}{4}\bigg(t^2-2\cdot 3t+9-9-72\bigg )\\ &= \frac{1}{4}\bigg((t-3)^2-81\bigg )\\ &= \frac{1}{4}\bigg((t-3)^2-9^2\bigg )\\ &= \frac{1}{4}\bigg((t-3)-9\bigg )\bigg((t-3)+9\bigg )\\ &= \frac{1}{4}\bigg(t-12\bigg )\bigg(t+6\bigg )\\ &= \frac{1}{4}\bigg(2x-12\bigg )\bigg(2x+6\bigg )\\ &= \bigg(x-6\bigg )\bigg(x+3\bigg )\\ \end{align}$$ This method is probably overkill, but it shines when used to derive the quadratic formula.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.