Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(x)=2 x^{4}-7 x^{3}-13 x^{2}+63 x-45$

The factors of constant term - 45 are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45

The factors of the coefficient of $x^{4}$ is 2 . Hence possible rational roots of $f(x)$ are

± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 45/2

Let, x – 1 = 0

=> x = 1

f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45

= 2 – 7 – 13 + 63 – 45

= 0

Let, x – 3 = 0

=> x = 3

$f(3)=2(3)^{4}-7(3)^{3}-13(3)^{2}+63(3)-45$

= 162 – 189 – 117 + 189 – 45

= 0

So, (x – 1) and (x – 3) are the roots of f(x)

$\Rightarrow x^{2}-4 x+3$ is the factor of $f(x)$

Divide $f(x)$ with $x^{2}-4 x+3$ to get other three factors

By long division,

$2 x^{2}+x-15$ $x^{2}-4 x+32 x^{4}-7 x^{3}-13 x^{2}+63 x-45$ $2 x^{4}-8 x^{3}+6 x^{2}$ $(-) \quad(+) \quad(-)$ $x^{3}-19 x^{2}+63 x$ $x^{3}-4 x^{2}+3 x$ $(-) \quad(+) \quad(-)$ $-15 x^{2}+60 x-45$ $-15 x^{2}+60 x-45$ $(+) \quad(-) \quad(+)$

$x^{3}-19 x^{2}+63 x$

$x^{3}-4 x^{2}+3 x$

$(-) \quad(+) \quad(-)$

$-15 x^{2}+60 x-45$

$-15 x^{2}+60 x-45$

(+) $\quad(-) \quad(+)$

0

$=>2 x^{4}-7 x^{3}-13 x^{2}+63 x-45=\left(x^{2}-4 x+3\right)\left(2 x^{2}+x-15\right)$

$=>2 x^{4}-7 x^{3}-13 x^{2}+63 x-45=(x-1)(x-3)\left(2 x^{2}+x-15\right)$

Now,

$2 x^{2}+x-15=2 x^{2}+6 x-5 x-15$

= 2x(x + 3) – 5 (x + 3)

= (2x – 5) (x + 3)

So, $2 x^{4}-7 x^{3}-13 x^{2}+63 x-45=(x-1)(x-3)(x+3)(2 x-5)$