1.

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases : (i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x+1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3

Answer»

(i) p(x) = 2x3 + x2 – 2x – 1 

g(x) = x + 1 

If x + 1 = 0, then x = -1 

p(x) = 2x3 + x2 – 2x – 1 

p(-1)= 2(-1)3 + (-1)2 -2(- 1) – 1 

= 2(-1) + (1) + 2 – 1 = -2 + 1 + 2 – 1 

P(-1)= 0 

Here, r(x) = p(a) = 0, 

∴ g(x) is the a factor f(x) 

(ii) p(x) = x3 + 3x2 + 3x + 1 

g(x) = x + 2 

If x + 2 = 0, then x = -2 

∴ p(x) = x3 + 3x2 + 3x + 1 

p(-2) = (-2)3 + 3(-2)2 + 3 (-2) + 1 

= -8 + 3(4) + 3(-2) + 1 

= -8 + 12 – 6 + 1 

= 13 – 24 

p(-2)= -11 

Here we have r(x) = p(a) =-11. 

Value of r(x) is not equal to zero. 

∴ g(x) is not a factor of f(x). 

(iii) p(x) = x3 – 4x2 + x + 6 

g(x) = x – 3 

Let, x – 3 = 0, then x = 3 

p(x) = x3 – 4x2 + x + 6 

p(3) = (3)2 – 4(3)2 + 3 + 6 

= 27 – 4(9) + 3 + 6 

= 27 – 36 + 3 + 6 = 36 – 36 

p(3) = 0 

Here, we have r(x) = p(a) = 0 

∴ (x – 3) is the factor of p(x).



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