Question

2x^(3)+9x^(2)+7x-6 (Factorise by Factor
Theorem
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Answer 1

Hi.......

Hope it helps you

Answer 2

Answer:

$2 x^{3}-9 x^{2}+7 x+6=(x-2)(x-3)(2 x+1)$

Step-by-step explanation:

Given: A polynomial $2x^{3} +9x^{2} +7x-6$.

We need to factorize the polynomial using factor theorem.

The factor theorem states that if f(x) is a polynomial of degree$n\geq 1$ and 'a' is any real number, then $(x - a)$ is a factor of f(x) if $f(a)=0$.

So, we need to find such $a$ for which $f(a)=0$.

For $\quad \mathrm{x}=0$

$p(0)=2(0)^{3}-9(0)^{2}+(7 \times 0)+6=0-0+0+6=6$

      $\neq 0$

So, $(x-0)$ is not factor of $f(x).$

For $x=2$

$\mathrm{p}(2)=2(2)^{3}-9(2)^{2}+(7 \times 2)+6\\=(2 \times 8)-(9 \times 4)+14+6\\=16-36+14+6\\=0$

Thus, $(x-2)$ is a factor of $f(x)$.

Now to find other factors we divide $f(x)$ by $(x-2)$.

BY division algorithm :

$2x^{3}+9x^{2} +7x-6=(x-2) (2 \mathrm{x}^{2}-5 \mathrm{x}-3)$.

Now factorize it as follows:

$\begin{aligned}&2 x^{2}-5 x-3 \\&=2 x^{2}-6 x+x-3 \\&=2 x(x-3)+1(x-3) \\&=(x-3)(2 x+1)\end{aligned}$

So, factor of $f(x)$ are $(x-2),(x-3),(2x+1)$.

Hence, $2 x^{3}-9 x^{2}+7 x+6=(x-2)(x-3)(2 x+1)$.

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