Question

It is given that $(x + 1)$ is a common factor of $(x^{3} + p x + q)$ and $(x^{3} + (1 - p) x^{2} + 19 x - 2 q)$ .

When

p

and

q

have some values, factorise

x^{3} + p x + q

and

x^{3} + (1 - p) x^{2} + 19 x - 2 q

completely.

Accepted Answer

Let $f (x) = (x^{3} + p x + q)$ , $g (x) = (x^{3} + (1 - p) x^{2} + 19 x - 2 q)$

Given that, $(x + 1)$ is a common factor of $f (x)$ and $g (x)$ .

We know that, if $(x - a)$ is a factor of $h (x)$ then, $h (a) = 0$ .

From factor theorem, $f (- 1) = 0$

$\Rightarrow {(- 1)}^{3} + p (- 1) + q = 0 \Rightarrow - 1 - p + q = 0 \Rightarrow q = p + 1 . . . . . . . . (i) e q .$

And, $g (- 1) = 0$

$\Rightarrow {(- 1)}^{3} + (1 - p) {(- 1)}^{2} + 19 (- 1) - 2 q = 0 \Rightarrow - 1 + 1 - p - 19 - 2 q = 0 \Rightarrow - p - 19 - 2 (p + 1) = 0 \Rightarrow - p - 19 - 2 p - 2 = 0 \Rightarrow - 3 p - 21 = 0 \Rightarrow 3 p = - 21 \Rightarrow p = - 7$

From $(i) e q .$ , $q = - 7 + 1$

$\Rightarrow q = - 6$

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

$f (- 1) = {(- 1)}^{3} - 7 (- 1) - 6 = - 1 + 7 - 6 = 0$ ,

So $(x + 1)$ is a factor of $f (x)$ .

$x^{3} - 7 x - 6$ can be written as,

$x^{3} - 7 x - 6 = (x + 1) (a x^{2} + b x + c) \Rightarrow x^{3} - 7 x - 6 = a x^{3} + (b + a) x^{2} + (c + b) x + c$

Comparing the coefficient of $x^{3}$ ,

$a = 1$ ,

Comparing the coefficient of $x^{2}$ ,

$\Rightarrow b + a = 0 \Rightarrow b + 1 = 0 \Rightarrow b = - 1$

Comparing the constant,

$\Rightarrow c = - 6$

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

$\Rightarrow f (x) = (x + 1) (x^{2} - 3 x + 2 x - 6) \Rightarrow f (x) = (x + 1) (x (x - 3) + 2 (x - 3)) \Rightarrow f (x) = (x + 1) (x - 3) (x + 2)$

Now, $g (x) = x^{3} + (1 - p) x^{2} + 19 x - 2 q = x^{3} + 8 x^{2} + 19 x + 12$

$g (- 1) = {(- 1)}^{3} + 8 {(- 1)}^{2} + 19 (- 1) + 12 = 0$ ,

So $(x + 1)$ is a factor of $g (x)$ .

$x^{3} + 8 x^{2} + 19 x + 12$ can be written as,

$x^{3} + 8 x^{2} + 19 x + 12 = (x + 1) (a x^{2} + b x + c) \Rightarrow x^{3} + 8 x^{2} + 19 x + 12 = a x^{3} + (b + a) x^{2} + (c + b) x + c$

Comparing the coefficient of $x^{3}$ ,

$a = 1$ ,

Comparing the coefficient of $x^{2}$ ,

$\Rightarrow b + a = 8 \Rightarrow b + 1 = 8 \Rightarrow b = 7$

Comparing the constant,

$\Rightarrow c = 12$

Hence, $g (x) = x^{3} + 8 x^{2} + 19 x + 12 = (x + 1) (x^{2} + 7 x + 12)$

$\Rightarrow g (x) = (x + 1) (x^{2} + 4 x + 3 x + 12) \Rightarrow g (x) = (x + 1) (x (x + 4) + 3 (x + 4)) \Rightarrow g (x) = (x + 1) (x + 4) (x + 3)$

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 6: EXERCISE 1E

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 6: EXERCISE 1E

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Algebra, Exercise 6: EXERCISE 1E with Hints & Solutions

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