Question

Check whether p(x) is a multiple of g(x) or not .
(i) p(x) = x3- 5x2+ 4x- 3; g(x) = x-2
(ii) p(x) = 2x3- 11x2- 4x+ 3; g(x) = 2x + 3

Answer 1

Hi pupil here's your answer ::

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please see the attached answer

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hope that it helps. . . . . . .

Answer 2

Answer:

Here remainder ≠ 0

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

Step-by-step explanation:

In the question

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

$g(x) = x\ -\ 2$

by factor theorem,

g(x) = 0

x - 2 = 0

x = 2

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

  = 8 - 5*4 + 8 - 3

 = 8 - 20 + 5

 = - 7

Here remainder ≠ 0

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

(ii) $p(x) = 2x^{3}\ -\ 11x^{2}\ -\ 4x\ +\ 3$

g(x) = 2x + 3

by factor theorem,

g(x) = 0

2x + 3 = 0

2x = - 3

x = -3/2

$p(-\frac{3}{2} ) = 2(-\frac{3}{2} )^{3}\ -\ 11(-\frac{3}{2} )^{2}\ -\ 4(-\frac{3}{2}) \ +\ 3$

$= 2\times(-\frac{27}{8} )\ -\ 11\times(\frac{9}{4})\ +\frac{12}{2} \ +\ 3$

$=-\frac{27}{4} \ -\frac{99}{4}\ +\frac{12}{2} \ +\ 3$

$=\frac{-27\ -\ 99\ +\ 24\ +\ 12}{4}$

$=\frac{-126\ +\ 36}{4}$

$=\frac{-90}{4}$

$=\frac{-45}{2}$

Here remainder ≠ 0

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