Question
Using the Remainder Theorem, factorise the expression $3x^3 + 10x^2 + x – 6$. Hence, solve the equation $3x^3 + 10x^2 + x – 6 = 0$.
✓
Answer
Left $f (x)=3 x^3+10 x^2+x-6$
$f$ or $x=-1$,
$f(x)=f(-1)=3(-1)^3+10(-1)^2+(-1)-6=-3+10-1-6=0$
Hence, $(x+1)$ is a factor of $f(x)$.
$\therefore 3 x^3+10 x^2+x-6=(x+1)\left(3 x^2+7 x-6\right)$
$=(x+1)\left(3 x^2+9 x-2 x-6\right)$
$=(x-1)[3 x(x-3)-2(x+3)]$
$=(x+1)(x+3)(3 x-2)$
now, $3 x^3+10 x^2+x-6=0$
$(x+1)(x+3)(3 x-2)=0$
$x=-1,-3, \frac{2}{3}$
$f$ or $x=-1$,
$f(x)=f(-1)=3(-1)^3+10(-1)^2+(-1)-6=-3+10-1-6=0$
Hence, $(x+1)$ is a factor of $f(x)$.
$\therefore 3 x^3+10 x^2+x-6=(x+1)\left(3 x^2+7 x-6\right)$
$=(x+1)\left(3 x^2+9 x-2 x-6\right)$
$=(x-1)[3 x(x-3)-2(x+3)]$
$=(x+1)(x+3)(3 x-2)$
now, $3 x^3+10 x^2+x-6=0$
$(x+1)(x+3)(3 x-2)=0$
$x=-1,-3, \frac{2}{3}$
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