Question
Using Remainder Theorem, factorise:
$x^3 + 10x^2 – 37x + 26$ completely
$x^3 + 10x^2 – 37x + 26$ completely
✓
Answer
By remainder theorem,
for $x=1,$ the value of the given expression is the remainder.
$x^3+10 x^2-37 x+26$
$=(1)^3+10(1)^2-37(1)+26$
$=1+10-37+26$
$=37-37$
$=0$
$x-1$ is a factor of $x^3+10 x^2-37 x+26$.
$\therefore x^3+10 x^2-37 x+26$
$=(x-1)\left(x^2+11 x-26\right)$
$=(x-1)\left(x^2+13 x-2 x-26\right)$
$=(x-1)[x(x+13)-2(x+13)]$
$=(x-1)(x+13)(x-2)$
$\therefore x^3+10 x^2-37+26=(x-1)(x+13)(x-2)$
for $x=1,$ the value of the given expression is the remainder.
$x^3+10 x^2-37 x+26$
$=(1)^3+10(1)^2-37(1)+26$
$=1+10-37+26$
$=37-37$
$=0$
$x-1$ is a factor of $x^3+10 x^2-37 x+26$.
$\therefore x^3+10 x^2-37 x+26$
$=(x-1)\left(x^2+11 x-26\right)$
$=(x-1)\left(x^2+13 x-2 x-26\right)$
$=(x-1)[x(x+13)-2(x+13)]$
$=(x-1)(x+13)(x-2)$
$\therefore x^3+10 x^2-37+26=(x-1)(x+13)(x-2)$
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