Question

Using factor theorem, factorize: $p\left(x\right)=2x^4-7x^3-13x^2+63x-45$

• A. $p\left(x\right)=(x-1)(x+3)(2x-5)$
• B. $p\left(x\right)=(x-1)(x-3)(x+3)(2x-5)$
• C. $p\left(x\right)=(x-1)(x-3)(2x-5)$
• D. $p\left(x\right)=(x+1)(x-3)(x+3)(2x+5)$

Answer 1

The correct option is B $p\left(x\right)=(x-1)(x-3)(x+3)(2x-5)$

$45\Rightarrow \pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 45$
If we put x = 1 in p(x)
$p\left(1\right)=2(1{)}^4-7(1{)}^3-13(1{)}^2+63(1)-45$
$p\left(1\right)=2-7-13+63-45=65-65=0$
$\therefore x=1$ or x - 1 is a factor of p(x).
Similarly if we put x = 3 in p(x)
$p\left(3\right)=2(3{)}^4-7(3{)}^3-13(3{)}^2+63(3)-45$
$p\left(3\right)=162-189-117+189-45=162-162=0$
Hence, x = 3 or (x - 3) = 0 is the factor of p(x).
$p\left(x\right)=2x^4-7x^3-13x^2+63x-45$
$\therefore p\left(x\right)=2x^3(x-1)-5x^2(x-1)-18x(x-1)+45(x-1)$
$\Rightarrow p\left(x\right)=(x-1)(2x^3-5x^2-18x+45)$
$\Rightarrow p\left(x\right)=(x-1)(2x^3-5x^2-18x+45)$
$\Rightarrow p\left(x\right)=(x-1)\left[2x^2\right(x-3)+x(x-3)-15(x-3\left)\right]$
$\Rightarrow p\left(x\right)=(x-1)(x-3)(2x^2+x-15)$
$\Rightarrow p\left(x\right)=(x-1)(x-3)(2x^2+6x-5x-15)$
$\Rightarrow p\left(x\right)=(x-1)(x-3)\left[2x\right(x+3)-5(x+3\left)\right]$
$\Rightarrow p\left(x\right)=(x-1)(x-3)(x+3)(2x-5).$