Question
$2x^4 - 7x^3 - 13x^2 + 63x - 45$
✓
Answer
Let $f(x)=2 x^4-7 x^3-13 x^2+63 x-45$ be the given polynomial.
Now, putting $x=1$, we get $f(1)=2(1)^4-7(1)^3-13(1)^2+63(1)-45$
$=2-7-13+63-45=65-65=0$
Therefore, $(x-1)$ is a factor of polynomial $f(x)$.
Now, $f(x)=2 x^3(x-1)-5 x^2(x-1)-18 x(x-1)+45(x-1)$
$=(x-1)\left(2 x^3-5 x^2-18 x+45\right)=(x-1) g(x) \ldots(1)$
$\text { Where } g(x)=2 x^3-5 x^2-18 x+45 \text { Putting } x=3 \text {, }$
$\text { we get: } g(3)=2(3)^3-5(3)^2-18(3)+45$
$=54-45-54+45=0$
Therefore, $(x-3)$ is the factor of $g(x)$.
Now, $g(x)=2 x^2(x-3)+x(x-3)-15(x-3)$
$=(x-3)\left(2 x^2+x-15\right)$
$=(x-3)\left(2 x^2+6 x-5 x-15\right)$
$=(x-3)(x+3)(2 x-5) \ldots(2)$
From equation (1) and (2),
we get: $f(x)=(x-1)(x-3)(x+3)(2 x-5)$
Hence, $(x-1),(x-3),(x+3)$ and $(2 x-5)$ are the factors of polynomial $f(x)$.
Now, putting $x=1$, we get $f(1)=2(1)^4-7(1)^3-13(1)^2+63(1)-45$
$=2-7-13+63-45=65-65=0$
Therefore, $(x-1)$ is a factor of polynomial $f(x)$.
Now, $f(x)=2 x^3(x-1)-5 x^2(x-1)-18 x(x-1)+45(x-1)$
$=(x-1)\left(2 x^3-5 x^2-18 x+45\right)=(x-1) g(x) \ldots(1)$
$\text { Where } g(x)=2 x^3-5 x^2-18 x+45 \text { Putting } x=3 \text {, }$
$\text { we get: } g(3)=2(3)^3-5(3)^2-18(3)+45$
$=54-45-54+45=0$
Therefore, $(x-3)$ is the factor of $g(x)$.
Now, $g(x)=2 x^2(x-3)+x(x-3)-15(x-3)$
$=(x-3)\left(2 x^2+x-15\right)$
$=(x-3)\left(2 x^2+6 x-5 x-15\right)$
$=(x-3)(x+3)(2 x-5) \ldots(2)$
From equation (1) and (2),
we get: $f(x)=(x-1)(x-3)(x+3)(2 x-5)$
Hence, $(x-1),(x-3),(x+3)$ and $(2 x-5)$ are the factors of polynomial $f(x)$.
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