Using factor theorem factorize $x^{3} - 6 x^{2} + 3 x + 10$ .

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Solution

For factorising the given equation, we should know at least one zero of this polynomial. Then we divide the polynomial by the factor to find the quotient and factor the quotient further to find other zeroes.
Let $x = 1$
$\Rightarrow 1^{3} - 6 {(1)}^{2} + 3 (1) + 10 \neq 0 \Rightarrow 1 - 6 + 3 + 10 \neq 0 \Rightarrow 8 \neq 0$
Let $x = 2$
$\Rightarrow 2^{3} - 6 {(2)}^{2} + 3 (2) + 10 = 0 \Rightarrow 8 - 24 + 6 + 10 = 0 \Rightarrow 0 = 0$
Therefore, $(x - 2)$ is a factor.
Other factor = $\frac{x^{3} - 6 x^{2} + 3 x + 10}{x - 2} = x^{2} - 4 x - 5 x - 2 x^{2} - 4 x - 5 x^{3} - 6 x^{2} + 3 x + 10 x^{3} - 2 x^{2} - + - 4 x^{2} + 3 x - 4 x^{2} + 8 x + - - 5 x + 10 - 5 x + 10 + - 0$
Now
Quotient = $x^{2} - 4 x - 5$
By Using the common factor theorem,
$\Rightarrow x^{2} + x - 5 x - 5 \Rightarrow x (x + 1) - 5 (x + 1) \Rightarrow (x + 1) (x - 5)$
To find zeroes
$(x + 1) (x - 5) = 0 \Rightarrow x = - 1, x = 5$
So,The zeroes of polynomial $x = - 1, 2, 5$ .
Therefore, $x^{3} - 6 x^{2} + 3 x + 10 = (x + 1) (x - 2) (x - 5)$ .

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