Factoring polynomials in specific format

Question

How do you write the polynomial $6x^2+7x-20$ on the form $a_n(x-r_1)(x-r_2) \cdots (x-r_n)$?

I cannot get it to factor correctly using common practices.

Answer 1

To factor the quadratic $ax^2+bx+c$, look for two numbers whose product is $ac$ and whose sum is $b$. In your case, the product and sum are $-120$ and $7$, respectively, so the numbers we're looking for are $15$ and $-8$.

We now rewrite the middle term (i.e., $bx$) using the two numbers we found, then take out common factors in pairs: $$ \begin{align} 6x^2+7x-20 &= 6x^2+15x-8x-20\\ &=3x(2x+5) - 4(2x+5)\\ &=(3x-4)(2x+5) \end{align} $$ We now factor out the leading coefficients from both terms: $$ \begin{align} (3x-4)(2x+5) &= 3(x-\frac43)\cdot2(x+\frac52)\\ &=6(x-\frac43)(x+\frac52) \end{align} $$

Answer 2

Since the polynomial \begin{align*} 6x^2+7x-20=6\left(x^2+\frac{7}{6}x-\frac{20}{6}\right)\tag{1} \end{align*} has degree $2$, we are looking for a representation \begin{align*} a_2(x-r_1)(x-r_2)\tag{2} \end{align*} Expanding (2) gives \begin{align*} a_2(x-r_1)(x-r_2)&=a_2\left(x^2-(r_1+r_2)x+r_1r_2\right)\tag{3} \end{align*} Comparing coefficients of (1) and (3) results in \begin{align*} a_2&=6\\ r_1+r_2&=-\frac{7}{6}\tag{4}\\ r_1\cdot r_2&=-\frac{20}{6} \end{align*} We observe $a_2=6$ and we could solve the system of the other two equations to find $r_1$ and $r_2$.

But we instead solve the quadratic equation

\begin{align*} 6x^2+7x-20=0 \end{align*} and get the two solutions \begin{align*} x_{1,2}&=\frac{1}{12}\left(-7\pm\sqrt{7^2-4\cdot6\cdot(-20)}\right)\\ &=\frac{1}{12}\left(-7\pm\sqrt{529}\right)\\ &=\frac{1}{12}(-7\pm23)\\ &= \begin{cases} +\frac{16}{12}=+\frac{4}{3}\\ -\frac{30}{12}=-\frac{5}{2} \end{cases} \end{align*}

So, we finally obtain \begin{align*} 6x^2+7x-20&=6\left(x-x_1\right)\left(x-x_2\right)\\ &=6\left(x-\frac{4}{3}\right)\left(x+\frac{5}{2}\right) \end{align*}

We can check this representation and find in accordance with (4) \begin{align*} x_1+x_2&=\frac{4}{3}-\frac{5}{2}=\frac{8}{6}-\frac{15}{6}=-\frac{7}{6}=r_1+r_2\\ x_1\cdot x_2&=\frac{4}{3}\cdot\left(-\frac{5}{2}\right)=-\frac{20}{6}=r_1\cdot r_2 \end{align*}

Answer 3

The AC-method reduces the problem to factoring a polynomial that is monic (lead coef $= 1).\,$

$$ 0 = 6f(x) = (6x)^2 + 7(6x) - 6(20) = X^2 + \color{#c00}7 X \color{#0a0}{-120},\ \ X = 6x\qquad $$

By Vieta, roots $\,a,b\,$ satisfy $\,a+b = -\color{#c00}7,\,$ and $\,ab = \color{#0a0}{-120} = -3\cdot 5\cdot 8. \,$ But $\,-3\cdot 5+ 8 = -7,\,$ which yields $\, 6f(x)= X^2+7X+120 = (X-8)(X+15) = (6x-8)(6x+15).\,$ Next cancel $\,6\,$ via: $ $ cancel $\,2\,$ from 1st, $\,3\,$ from 2nd factor, $ $ so $\, f(x) = (3x-4)(2x+5).\,$

Remark $\ $ Such cancellation will always be possible by Gauss's Lemma (see the linked post).

More mechanically, one could simply apply the quadratic formula. But tricks like the above are often quicker for small numbers.

Answer 4

Given the factorization $6x^2-7x-20=(2x+5)(3x-4)$ you can then bring the coefficients of $x$ to the front, getting $6(x+\frac 52)(x-\frac 43)$ as required.