Question

If the polynomial $2 x^{3} + a x^{2} + 7 x - 6$ is exactly divisible by $(2 x - 1)$ , then find the value of $a$ . Hence, factorise the polynomial.

Accepted Answer

We have $p (x) = 2 x^{3} + a x^{2} + 7 x - 6$

Factor Theorem: If $p (x)$ is a polynomial of degree $\geq 1$ and $α$ is real number, then $x - α$ is a factor of $p (x)$ if $p (α) = 0$ and conversely, $p (α) = 0$ if $x - α$ is a factor of $p (x)$ .

So, if $2 x^{3} + a x^{2} + 7 x - 6$ is divisible by $(2 x - 1)$ , then $(2 x - 1)$ is a factor of $2 x^{3} + a x^{2} + 7 x - 6$ .

By factor theorem, we have $p (\frac{1}{2}) = 0$

$\Rightarrow 2 {(\frac{1}{2})}^{3} + a {(\frac{1}{2})}^{2} + 7 (\frac{1}{2}) - 6 = 0$

$\Rightarrow \frac{1}{4} + \frac{a}{4} + \frac{7}{2} - 6 = 0$

$\Rightarrow \frac{1 + a + 14}{4} = 6$

$\Rightarrow a + 15 = 24$

$\Rightarrow a = 24 - 15 = 9$

Now, $p (x) = 2 x^{3} + 9 x^{2} + 7 x - 6$

To find the other factors of $2 x^{3} + 9 x^{2} + 7 x - 6$ , we need to divide $p (x)$ by $(2 x - 1)$ .

Perform $(2 x^{3} + 9 x^{2} + 7 x - 6) \div (2 x - 1)$ as follows.

$\begin{matrix} x^{2} & + & 5 x & + & 6 \\ 2 x - 1 & 2 x^{3} & + & 9 x^{2} & + & 7 x & - & 6 \\ 2 x^{3} & - & x^{2} \\ 10 x^{2} & + & 7 x \\ 10 x^{2} & - & 5 x \\ 12 x & - & 6 \\ 12 x & - & 6 \\ 0 \end{matrix}$

$(2 x^{3} + 9 x^{2} + 7 x - 6) \div (2 x - 1) = x^{2} + 5 x + 6$

We know that $x^{2} + (a + b) x + a b = (x + a) (x + b)$

$x^{2} + 5 x + 6 = x^{2} + 2 x + 3 x + 6$

$x^{2} + 5 x + 6 = x (x + 2) + 3 (x + 2)$

$x^{2} + 5 x + 6 = (x + 2) (x + 3)$

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

Therefore, $a = 9$ and the factors of $2 x^{3} + a x^{2} + 7 x - 6$ are $(2 x - 1), (x + 2), (x + 3)$ .

M L Aggarwal Solutions for Chapter: Polynomials, Exercise 6: Exercise 2.6

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Polynomials, Exercise 6: Exercise 2.6

Questions from M L Aggarwal Solutions for Chapter: Polynomials, Exercise 6: Exercise 2.6 with Hints & Solutions

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