Factor $x^2+10x+21$

Question

How can we factor $x^2+10x+21$? I'm a bit confused. Is the answer $(x+7)(x+3)$? I know that $7+3=10$ and $7\times 3=21$.

How can we factor $x^2+5x-24$? How about $x^2-x-20$? Why do people say $abc$ or $eb+ec$? I don't get it!

Answer 1

Assuming first problem written incorrectly.

$$(x+3)(x+7)$$

$$(x-3)(x+8)$$

$$(x+4)(x-5)$$

Answer 2

1.$$x^2 + 10x + 21=(x^2+7x)+(3x+21)=x(x+7)+3(x+7)=(x+3)(x+7)$$ 2.$$x^2 + 5x - 24=x^2+5x-15-9=(x^2-3^2)+(5x-15)=$$ $$=(x+3)(x-3)+5(x-3)=(x-3)(x+8)$$

3.$$x^2-x-20=(x^2-5x)+(4x-20)=x(x-5)+4(x-5)=(x-5)(x+4)$$

Answer 3

Let me answer your question generally, since all three quadratics have the same form.

Suppose that you wish to factorize $x^2+px+q$. If you can find $a,b$ such that $a+b=p$ and $ab=q$, then we have $$\begin{align}x^2+px+q &= x^2+(a+b)x+ab\\ &= x^2+ax+bx+ab\\ &= (x^2+ax)+(bx+ab)\\ &= (x+a)x+(x+a)b\\ &= (x+a)(x+b).\end{align}$$

For example, consider your third quadratic, $x^2-x-20=x^2+(-1)x+(-20)$. Putting $a=4$ and $b=-5$, we have $a+b=-1=p$ and $ab=-20=q$, so we have $$x^2-x-20=(x+4)(x+-5)=(x+4)(x-5),$$ as you can (and should) check. If we're being careful, we'll actually go through the steps, so $$\begin{align}x^2-x-20 &= x^2+(4-5)x-20\\ &= x^2+4x-5x-20\\ &= (x^2+4x)+(-5x-20)\\ &= (x+4)x+(x+4)(-5)\\ &= (x+4)(x-5).\end{align}$$

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Sometimes, though, finding such $a,b$ is an exercise in futility--we can always go by trial and error among integer factors of $q$, but sometimes that won't work. In such a situation, we can use a method called completing the square. We can finagle things as follows, using the fact that $(y+z)(y-z)=y^2-z^2$ for all complex $y,z$: $$\begin{align}x^2+px+q &= (x+p)x+q\\ &= \left(x+\frac{p}2+\frac{p}2\right)\left(x+\frac{p}2-\frac{p}2\right)+q\\ &= \left(x+\frac{p}2\right)^2-\left(\frac{p}2\right)^2+q\\ &= \left(x+\frac{p}2\right)^2-\frac{p^2}4+q\\ &= \left(x+\frac{p}2\right)^2-\frac{p^2-4q}4\\ &= \left(x+\frac{p}2\right)^2-\left(\sqrt{\frac{p^2-4q}4}\right)^2\\ &= \left(x+\frac{p}2\right)^2-\left(\frac{\sqrt{p^2-4q}}2\right)^2\\ &= \left(\left(x+\frac{p}2\right)+\frac{\sqrt{p^2-4q}}2\right)\left(\left(x+\frac{p}2\right)-\frac{\sqrt{p^2-4q}}2\right)\\ &= \left(x+\frac{p+\sqrt{p^2-4q}}2\right)\left(x+\frac{p-\sqrt{p^2-4q}}2\right).\end{align}$$

We can also use the quadratic formula (which is just a shortcut to completion of the square, really), together with the fact that $x_0$ is a root of the quadratic if and only if $x-x_0$ is a factor. The quadratic formula for $x^2+px+q$ gives us the roots $$\frac{-p\pm\sqrt{p^2-4q}}2,$$ so the factors are $$x+\frac{p\mp\sqrt{p^2-4q}}2,$$ as we saw above.

Answer 4

There are various ways of looking at this kind of problem. One is to see that:$$(x+a)(x+b)=x^2+(a+b)x+ab$$

and identify $a$ and $b$ from the coefficients by examination. This generalises so that, for example:$$(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(bc+ac+ab)x+abc$$ - so the coefficients appear as elementary symmetric functions, and this works in higher degrees too. In fact we normally work with $p(x)=(x-a)(x-b)(x-c)$ because if we want to solve $p(x)=0$ and we postulate $p(a)=0$ we find that $0=p(x)=p(x)-p(a)=(x-a)q(x)$ where $q(x)$ is a polynomial of lower degree. It turns out to be easier to work with the roots of an equation, rather than their negatives, and we are then working with elementary symmetric polynomials in the roots. So we would conventionally write:$$(x-a)(x-b)=x^2-(a+b)x+ab$$

Another, more general , way of proceeding is to note that if we have $y^2=a$ then we can write $y=\pm \sqrt a$. So we rewrite $$x^2+px+q=0$$ as $$x^2+px+\frac{p^2}4=\frac{p^2}4-q$$ which is $$\left(x+\frac p2\right)^2=\frac{p^2}4-q$$ which is of the requisite form.