How do you factor completely 6x^2+7x+2?

Mar 28, 2018

$\left(2 x + 1\right) \left(3 x + 2\right)$

Explanation:

$\text{for a quadratic in "color(blue)"standard form}$

•color(white)(x)ax^2+bx+c color(white)(x);a!=0

$6 {x}^{2} + 7 x + 2 \leftarrow \text{is in standard form}$

$\text{with "a=6, b=7" and } c = 2$

$\textcolor{b l u e}{\text{to factorise}}$

$\text{consider the factors of the product ac which sum to b}$

$\Rightarrow a c = 6 \times 2 = 12 \text{ and + 4, + 3 sum to + 7}$

$6 {x}^{2} + 3 x + 4 x + 2 \leftarrow \textcolor{b l u e}{\text{split middle term}}$

$= \textcolor{red}{3 x} \left(2 x + 1\right) \textcolor{red}{+ 2} \left(2 x + 1\right) \leftarrow \textcolor{b l u e}{\text{factor in groups}}$

$\text{take out a common factor } \left(2 x + 1\right)$

$= \left(2 x + 1\right) \left(\textcolor{red}{3 x + 2}\right)$

$\Rightarrow 6 {x}^{2} + 7 x + 2 = \left(2 x + 1\right) \left(3 x + 2\right)$

Mar 28, 2018

$\left(2 x + 1\right) \left(3 x + 2\right)$

Explanation:

By sum & product

= $6 {x}^{2} + 4 x + 3 x + 2$

= $2 x \left(3 x + 2\right) + 1 \left(3 x + 2\right)$

= $\left(2 x + 1\right) \left(3 x + 2\right)$

Hope this helps!