# a more scientific way to factor

I used trial and error to factorise the following expression:

${9p^2 + 18p -16}$

I went through a number of different possibilities until I discovered the answer:

${(3p - 2)(3p + 8)}$

Is there an algorithm I could have used, using the coefficients or something that would be better?

• Use middle term breaking.. – tatan Sep 3 '15 at 9:04
• You can find the solutions of $9p^2 + 18p -16 =0$, which are $\frac{2}{3}$ and $\frac{-8}{3}$. Now, do you see the method? – GAVD Sep 3 '15 at 9:07
• Quadratic factorization – Peđa Terzić Sep 3 '15 at 9:11

Suppose $(rx + s)(tx + u) = ax^2 + bx + c$. Then \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\ & = \color{blue}{rt}x^2 + (\color{green}{ru} + \color{green}{st})x + \color{blue}{su} \end{align*} Matching coefficients, we see that $a = rt$, $b = ru + st$, and $c = su$.

Observe that the product of the two coefficients that sum to the coefficient of the linear term is equal to the product of the coefficients of the quadratic and constant terms, that is, $$(\color{green}{ru})(\color{green}{st}) = (\color{blue}{rt})(\color{blue}{su}) = rstu$$

Thus, if a quadratic polynomial admits a factorization with respect to the rational numbers, the linear term splits into two numbers with product $ac$ and sum $b$.

In the example $9p^2 + 18p - 16$, we must find two numbers with product $9 \cdot -16 = -144$ and sum $18$. They are $-6$ and $24$. Hence, \begin{align*} 9p^2 + 18p - 16 & = 9p^2 - 6p + 24p - 16 && \text{split the linear term}\\ & = 3p(3p - 2) + 8(3p - 2) && \text{factor by grouping}\\ & = (3p + 8)(3p - 2) && \text{extract the common factor} \end{align*}

You may know the quadratic formula which states that the roots of $$ax^2+bx+c = 0$$ are given by $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Something you may not know is that we can rewrite the original polynomial as $a(x-x_1)(x-x_2)$ where $x_1,x_2$ are the roots. Hence a factorization is \begin{align} ax^2+bx+c &= a\left(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\right). \end{align}

• what you gona do if the equation has 4th power? – Salihcyilmaz Sep 3 '15 at 15:55
• @Salihcyilmaz Well, if you are trying to factor a 4th degree polynomial it gets much more difficult. There is actually a closed form formula, but it is quite hairy. There is no general easy way for higher order polynomials. However, whenever one can find one root, the problem can be reduced by polynomial division. – Eff Sep 3 '15 at 16:06
• Well, i didn't now that there was a closed form for 4th degree polynomial, nice to know :) What i simply referred was higher order polynomials as you stated. Then we can use some numeric methods, maybe newtons method. – Salihcyilmaz Sep 3 '15 at 16:10

The most natural form for a quadratic expression is $y = a(x-b)^2+c$ where a is non-zero and b, c can be any numbers. This is the "completed square" form and any quadratic can be put into this form. In your case, start by factoring out the 9 in the first two terms: $$y = 9(p^2+2p) - 16$$ In general, this step might create the coefficient b as a fraction. Now (and this is the important step), $p^2+2p$ can be written as $(p+1)^2 - 1^2$. The coefficient of p is 2, half of this is 1, and you write down $(p+1)^2 - 1^2$. In a similar way $x^2+4x=(x+2)^2 - 2^2$, $x^2-4x=(x-2)^2 -(-2)^2$, $x^2+5x=(x+\frac{5}{2})^2 - (\frac{5}{2})^2$ and so on. Returning to your problem we now have $$y = 9((p+1)^2 - 1^2) - 16=9((p+1)^2 - 1) - 16=9(p+1)^2 - 9 - 16=9(p+1)^2 - 25$$ In this form, since $(p+1)^2$ is always positive, we can see that y has a minimum possible value of -25 and this minimum is reached when p = -1. To factorize the quadratic we use the "difference of 2 squares" formula, which is $X^2-Y^2 = (X-Y)(X+Y)$. For your quadratic $y = 9(p+1)^2 - 25$ we have $X=3(p+1)$ and $Y = 5$. Then $$y = [3(p+1)-5][3(p+1)+5] = [3p-2][3p+8]$$

This process, with a bit of practice, is pretty quick and not only lets you factorize a quadratic expression and solve a quadratic equation but also tells you the nature of the roots (c<0 gives real roots, c = 0 a single root and c>0 imaginary roots) and the location of the min(c<0 when a is positive) or the max (c>0 when a is positive) of the quadratic

Use middle term breaking. Let, $p(x)=x^2+bx+c$

Let, the factors be $(x+p)(x+q).$

So,we get $x^2+bx+c=(x+p)(x+q)=x^2+(p+q)x+pq.$

Comparing we get $b=(p+q)$ and $c=pq.$

So,you have to find numbers $p$ and $q$ such that $pq=c$ (constant term) and $p+q=b$.

If the question is of the form $ax^2+bx+c$ (like yours) $pq≠c$ but $ac$.(Try to prove it,why?)

Now, coming back to your problem..

$9p^2+18p−16$

So,$c=-16\times 9=-144$ which is to be expressed in $pq$ form.Prime factors of $-144$ are $2,2,2,2,3,3$.Now this is to be arranged in the form such that $p+q=18$ and $pq=-144$.We see that $24$ and $-6$ are the required $p$ and $q$ $[24+(-6)=18]$ and $(24\times-6=-96).$

So,we can write,

$9p^2+18p−16$

=$9p^2+24p-6p−16$

=$3p(3p+8)-2(3p+8)$

=$(3p-2)(3p+8)$

Factor both coefficients of the x squared term and the constant. Now combine these so that they multiply to the constant and add or subtract to the linear term. It the second sing is positive then both binomials agree with the sign for the linear term. If the constant term is negative the linear sign tells you the sign of the bigger combination from from the inner terms and the outer terms.

Use the rational root theorem. The coefficients have no common factor, so if there is a zero $r = a/b$, then $a$ divides the constant term and $b$ the coefficient of the leading term. This cuts down the possibilities. By Descartes' rule of signs there is one real positive zero, finding that one gives the other by division.

In any case, this is a quadratic, so the quadratic formula tells all:

$$r = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 9 \cdot (-16)}}{2 \cdot 9}$$

thus $$\left\{-\frac{8}{3}, \frac{2}{3} \right\}$$