Factor the expression (6x^2-3x+2)^2-1

 Exercise

$\left(6x^2-3x+2\right)^2-1$

 

Use the complete the square method to factor the trinomial of the form $ax^2+bx+c$. Take common factor $a$ ($6$) to all terms

$\left(6\left(x^2-\frac{1}{2}x+\frac{1}{3}\right)\right)^2-1$

 

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

$\left(6\left(x^2-\frac{1}{2}x+\frac{1}{3}+\frac{1}{16}-\frac{1}{16}\right)\right)^2-1$

 

Factor the perfect square trinomial $x^2+-\frac{1}{2}x+\frac{1}{16}$

$\left(6\left(\left(x- \frac{1}{4}\right)^2+\frac{1}{3}-\frac{1}{16}\right)\right)^2-1$

 

Multiply the fraction and term in $- \frac{1}{4}$

$\left(6\left(\left(x-\frac{1}{4}\right)^2+\frac{1}{3}-\frac{1}{16}\right)\right)^2-1$

$\left(6\left(\left(x-\frac{1}{4}\right)^2+\frac{1}{3}-\frac{1}{16}\right)\right)^2-1$

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