If $p (x) = x^{3} - 5 x^{2} + 4 x - 3 and g (x) = x - 2$ , show that p(x) is not a multiple of g(x).

Open in App

Solution

$p (x) = x^{3} - 5 x^{2} + 4 x - 3$

$g (x) = x - 2$

Putting x = 2 in p(x), we get

$p (2) = 2^{3} - 5 \times 2^{2} + 4 \times 2 - 3 = 8 - 20 + 8 - 3 = - 7 \neq 0$

Therefore, by factor theorem, (x − 2) is not a factor of p(x).

Hence, p(x) is not a multiple of g(x).

Suggest Corrections

Similar questions

p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2

p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1

If $p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$ and $g (x) = 2 x + 1$ , show that g(x) is not multiple of p(x).

p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1

Check whether p(x) = $x^{3}$ + 5 $x^{2}$ + 2x - 5 is a multiple of g(x) = x + 2.

Join BYJU'S Learning Program