Using factor theorem, factorize of the polynomials:
$2 x^{4}-7 x^{3}-13 x^{2}+63 x-45$
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, $2 x^{4}-7 x^{3}-13 x^{2}+63 x-45=(x-1)(x-3)(x+3)(2 x-5)$
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