# USING FACTOR THEOREM TO SOLVE FOR UNKNOWN COEFFICIENT

Factor Theorem :

Let P(x) be a polynomial.

If (x - k) be a factor of P(x), then

P(k) = 0

Note :

If P(k) = 0, then (x - k) is not a factor of P(x).

## Using Factor Theorem to Solve for Unknown Coefficient

Problem 1 :

Find the value of k, if (x - 3) is a factor of

f(x) = x3 - 11x + k

Solution :

Given : (x - 3) is a factor of f(x).

By Factor Theorem,

f(3) = 0

33 - 11(3) + k = 0

27 - 33 + k = 0

-6 + k = 0

k = 6

Problem 2 :

Find the value of k, if (x + 1) is a factor of

f(x) = kx5 - 121x3 - 15x2 - 25

Solution :

Given : (x + 1) is a factor of f(x).

By Factor Theorem,

f(-1) = 0

k(-1)5 - 121(-1)3 - 15(-1)2 - 25 = 0

k(-1) - 121(-1) - 15(1) - 25 = 0

-k + 121 - 15 - 25 = 0

-k + 81 = 0

k = 81

Problem 3 :

What is the value of k, if (x + 2) is a factor of

f(x) = -(x3 + 3x2) - 4(x - a)

Solution :

Given : (x + 2) is a factor of f(x).

By Factor Theorem,

f(-2) = 0

-[(-2)3 + 3(-2)2] - 4(-2 - a) = 0

-[-8 + 3(4)] + 8 + 4a = 0

-[-8 + 12] + 8 + 4a = 0

-4 + 8 + 4a = 0

4 + 4a = 0

4a = -4

a = -1

Problem 4 :

If (x - 2) is a factor of polynomial

P(x) = a(x3 - 2x) + b(x2 - 5),

which of the following must be true?

(A)  a + b = 0

(B)  2a - b = 0

(C)  2a + b = 0

(D)  4a - b = 0

Solution :

Given : (x - 2) is a factor of f(x).

By Factor Theorem,

f(2) = 0

a[23 - 2(2)] + b(22 - 5) = 0

a(8 - 4) + b(4 - 5) = 0

a(4) + b(-1) = 0

4a - b = 0

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