Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 69 + 11x – x² + x³, g(x) = x + 3

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Solution

$p (x) = 69 + 11 x - x^{2} + x^{3}$

$g (x) = x + 3$

Putting x = −3 in p(x), we get

$p (- 3) = 69 + 11 \times (- 3) - {(- 3)}^{2} + {(- 3)}^{3} = 69 - 33 - 9 - 27 = 0$

Therefore, by factor theorem, (x + 3) is a factor of p(x).

Hence, g(x) is a factor of p(x).

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Using factor theorem, show that g(x) is a factor of p(x), when
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Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
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(ii) p(x)= $x^{3}$ +3 $x^{2}$ +3x+1,g(x)=x+2
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Q. Question 2
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1

(i i) p (x) = x^{3} + 3 x^{2} + 3 x + 1, g (x) = x + 2

(i i i) p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3

Show that (x+3) is a factor of p(x) =69+11x-x²+x³.

Use the factor theorem to determine and state whether

g (x)

is a factor of

p (x)

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

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Using factor theorem, show that g(x) is a factor of p(x), when p(x) = 69 + 11x – x2 + x3, g(x) = x + 3

px=69+11x-x2+x3 gx=x+3 Putting x = −3 in p(x), we get p-3=69+11×-3--32+-33=69-33-9-27=0 Therefore, by factor theorem, (x + 3) is a factor of p(x). Hence, g(x) is a factor of p(x).

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 69 + 11x – x² + x³, g(x) = x + 3

$p (x) = 69 + 11 x - x^{2} + x^{3}$

$g (x) = x + 3$

Putting x = −3 in p(x), we get

$p (- 3) = 69 + 11 \times (- 3) - {(- 3)}^{2} + {(- 3)}^{3} = 69 - 33 - 9 - 27 = 0$

Therefore, by factor theorem, (x + 3) is a factor of p(x).

Hence, g(x) is a factor of p(x).