Question

Factor the trinomial

$6x^2+11x-21$

Short answer

The solution$6x^2+11x-21=\left(6x-7\right)\left(x+3\right)$

Step-by-step solution

Step 1. Apply the concept of factoring trinomials.

To factor a trinomial of the form $a{x}^2+bx+c;a\ne 0$, we note that

$a{x}^2+bx+c=\left(px+r\right)\left(qx+s\right)=pq{x}^2+\left(ps+qr\right)x+rs$

So, we need to choose numbers p, q, r and s so that $pq=a$,$ps+qr=b$ and $rs=c$.

Step 2. Analyze the trinomial.

Given$6x^2+11x-21$

This is of the form $a{x}^2+bx+c$where$a=6$$b=11$ and$c=-21$

The factorization of the trinomial will be$\left(px+r\right)\left(qx+s\right)=pq{x}^2+\left(ps+qr\right)x+rs$

So, we need to choose numbers r and s so that $pq=a$,$ps+qr=b$ and $rs=c$.

Step 3. Find the integers to factor the trinomial.

We need to choose integers r and s so that $pq=6$,$ps+qr=11$ and $rs=-21$.

By trial and error, we find that the two integers are$p=6,q=1,r=-7,s=3$

$pq=6\cdot 1=6$

$ps+qr=18-7=11$

$rs=-7\cdot 3=-21$

Step 4. Factor the trinomial.

The factorization of the trinomial is

$\begin{array}{l}6x^2+11x-21=\left(px+r\right)\left(qx+s\right)\\ \Rightarrow 6x^2+11x-21=\left(6x-7\right)\left(x+3\right)\end{array}$