Mathematics
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
Polynomials
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Answer
(i) p(x) = 2x3 + x2 - 2x - 1
g(x) = x + 1
⇒ x + 1 = 0
⇒ x = -1
Putting x = -1 we get,
p(-1) = 2(-1)3 + (-1)2 - 2(-1) - 1
= -2 + 1 + 2 -1
= 0
Remainder is zero (0), so g(x) is factor of p(x).
(ii) p(x) = x3 + 3x2 + 3x + 1
g(x) = x + 2
⇒ x + 2 = 0
⇒ x = -2
Putting x = -2 we get,
p(-2) = (-2)3 + 3(-2)2 + 3(-2) + 1
= -8 + 3(4) - 6 + 1
= -8 + 12 - 6 + 1
= 13 - 14
= -1
Remainder is not zero (0), so g(x) is not a factor of p(x).
(iii) p(x) = x3 - 4x2 + x + 6
g(x) = x - 3
⇒ x - 3 = 0
⇒ x = 3
Putting x = -3 we get,
p(3) = (3)3 - 4(3)2 + 3 + 6
= 27 - 4(9) + 9
= 27 - 36 + 9
= 36 - 36
= 0
Remainder is zero (0), so g(x) is a factor of p(x).
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Related Questions
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x - 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x - 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Determine which of the following polynomials has (x + 1) a factor :
(i) x3 + x2 + x + 1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 - x2 - (2 + )x +
Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x2 + x + k
(ii) p(x) = 2x2 + kx +
(iii) p(x) = kx2 - x + 1
(iv) p(x) = kx2 - 3x + k
Factorise :
(i) 12x2 - 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x - 6
(iv) 3x2 - x - 4