# How do you factor 3t^3-7t^2-3t+7?

Jul 19, 2017

$\left(3 t - 7\right) \left(t - 1\right) \left(t + 1\right)$

#### Explanation:

$\text{split the expression into 2 groups}$

$\left(3 {t}^{3} - 7 {t}^{2}\right) + \left(- 3 t + 7\right)$

$\text{factorise each pair}$

$\textcolor{red}{{t}^{2}} \left(3 t - 7\right) \textcolor{red}{- 1} \left(3 t - 7\right)$

$\text{factor out } \left(3 t - 7\right)$

$\left(3 t - 7\right) \left(\textcolor{red}{{t}^{2} - 1}\right)$

${t}^{2} - 1 \text{ is a "color(blue)"difference of squares}$

$\left(3 t - 7\right) \left(t - 1\right) \left(t + 1\right) \leftarrow \text{ factors of difference of squares}$

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

Jul 19, 2017

$3 {t}^{3} - 7 {t}^{2} - 3 t + 7 = \left(t - 1\right) \left(t + 1\right) \left(3 t - 7\right)$

#### Explanation:

let

$f \left(t\right) = 3 {t}^{3} - 7 {t}^{2} - 3 t + 7$

All cubics have at least one real root, so using the factor theorem we will try to find one of teh roots.

$f \left(1\right) = 3 \times 1 - 7 \times 1 - 3 \times 1 = 7 = 3 - 7 - 3 + 7 = 0$

so$\text{ "t=1" is root "=>(t-1)" is a factor}$

$\therefore f \left(t\right) = 3 {t}^{3} - 7 t - 3 t + 7 = \left(t - 1\right) \left(a {t}^{2} + b t + c\right)$

from comparing coefficients

$\text{coeff. } {t}^{2}$

$L H S = 3$

RHS=a#

$\implies a = 3$

so we now have:

$3 {t}^{3} - 7 {t}^{2} - 3 t + 7 = \left(t - 1\right) \left(3 {t}^{2} + b t + c\right)$

compare the constant term

$L H S = + 7$

$R H S = - 1 \times c = - c$

$\therefore c = - 7$

$3 {t}^{3} - 7 {t}^{2} - 3 t + 7 = \left(t - 1\right) \left(3 {t}^{2} + b t - 7\right)$

compare coeff. $\text{ } {t}^{2}$

$L H S = - 7$

$R H S = - 3 + b$

$b - 3 = - 7 \implies b = - 4$

$3 {t}^{3} - 7 {t}^{2} - 3 t + 7 = \left(t - 1\right) \left(3 {t}^{2} - 4 t - 7\right)$

the question now is: can we factorise the quadratic?

$3 {t}^{2} - 4 t - 7$

$3 \times - 7 = - 21$

factors of$- 21$ that add to$- 4 \text{ }$are $- 7 \text{ " & " } + 3$

can be factorised

replacing the middle term with these two factors we proceed as follows:

$3 {t}^{2} - 4 t - 7 = 3 {t}^{2} - 7 t + 3 t - 7$

$= \left(3 {t}^{2} - 73 t - 7\right) + \left(3 t - 7\right)$

$t \left(3 t - 7\right) + \left(3 t - 7\right)$

$\left(3 t - 7\right) \left(t + 1\right)$

so the final factorisation

$3 {t}^{3} - 7 {t}^{2} - 3 t + 7 = \left(t - 1\right) \left(3 t - 7\right) \left(t + 1\right)$

$3 {t}^{3} - 7 {t}^{2} - 3 t + 7 = \left(t - 1\right) \left(t + 1\right) \left(3 t - 7\right)$