Question

Factorise the following polynomial using synthetic division:

$x^{3} - 3 x^{2} - 10 x + 24$

Accepted Answer

Factorisation using synthetic division method that helps to factorise a cubic polynomial into linear factors. If we identify one linear factor of cubic polynomial $x^{3} - 3 x^{2} - 10 x + 24$ then using synthetic division we can get the quadratic factors of $p (x)$ . Further if possible one can factorise the quadratic factor into linear factors.

Let $p (x) = x^{3} - 3 x^{2} - 10 x + 24$

$x - 1$ is a factor of $p (x)$ if and only if the sum of coefficients of $p (x)$ is $0$ .

$(x + 1)$ is a factor of $p (x)$ if and only if the sum of the coefficients of even power of $x$ , including constant is equal to the sum of the coefficients of odd power of $x$ .

Sum of all the co-efficients of given polynomial $= 1 - 3 - 10 + 24 = 25 - 13 = 12 \neq 0$

Hence, $(x - 1)$ is not a factor.

Sum of co-efficient of even powers with constant $= - 3 + 24 = 21$
Sum of co-efficients of odd powers $= 1 - 10 = - 9$
$21 \neq - 9$

Hence, $(x + 1)$ is not a factor.

By replacing $x$ $= 2$ , we get

$p (2) = 2^{3} - 3 (2^{2}) - 10 \times 2 + 24 = 8 - 12 - 20 + 24$

$= 32 - 32 = 0$

Therefore, $(x - 2)$ is factor

Thus, $(x - 2) (x + 3) (x - 4)$ are the factors of the given polynomial.

Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 13: Exercise

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 13: Exercise

Questions from Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 13: Exercise with Hints & Solutions

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