Use the factor theorem to factorise $a^{3} - 3 a^{2} + 4$ completely.

Open in App

Solution

Let p(a) = $a^{3} - a x^{2} + 4$
Using factor theorem,
Find the value of $p (- 1) = 0$
$p (a) = a^{3} - 3 a^{2} + 4$ ........(Substitute $a = - 1$ )
$p (- 1) = - 1^{3} - 3 (- 1)^{2} + 4$
$0 = - 1 - 3 + 4$
$0 = 0$
Therefore, $a + 1$ is a factor of $p (a)$ .
So, $p (a) = (a + 1) (a^{2} - 4 x + 4)$
$=$ $(a + 1) (a - 2) (a - 2)$
$=$ $(a + 1) (a - 2)^{2}$

Suggest Corrections

Similar questions

Use factor theorem to factorise completely: ³ + ² – 4 – 4.

Using factor theorem factorise the expression completely: 3³ + 2² – 19 + 6

3 x^{3} + 2 x^{2} - 19 x + 6

x^{4} - 5 x^{2} + 4

Using factor theorem show that ( – 3) is a factor of ³ – 7² + 15 – 9. Hence factorise the expression completely.

Join BYJU'S Learning Program