# REMAINDER THEOREM AND FACTOR THEOREM WORKSHEET

Problem 1 :

f(x) = x3 + 3x2 + 3x + 1

In the polynomial function f(x) defined above, using Remainder Theorem, find the remainder when f(x) is divided by (x + 1).

Problem 2 :

f(x) = x3 - 3x + 1

In the polynomial function f(x) defined above, using Remainder Theorem, find the remainder when f(x) is divided by (2 - 3x).

Problem 3 :

2x4 + 3x3 + 2kx2 + 3x + 6

Find the value of k for which the polynomial above is divisible by (x + 2).

Problem 4 :

x3 - 4x2 - 2x + 20

Using Factor Theorem, show that (x + 2) is a factor of the polynomial above.

Problem 5 :

Is (3x - 2) a factor of (3x3 + x2 - 20x + 12) ?

Problem 6 :

2x3 - 6x2 + mx + 4

Find the value of m, if (x - 2) is a factor of the polynomial above.

Problem 7 :

p(x) = 4x3 - kx + k

If (x + 1) divides p(x) without remainder, find the value of k.

Problem 8 :

p(x) = (3x2 - 5)(x + m) - 20

In the polynomial p(x) defined above, k is a constant. If x is a factor of p(x), what is the value of m? f(x) = x3 + 3x2 + 3x + 1

Equate the divisor (x + 1) to zero and solve for x.

x + 1 = 0

x = -1

By Remainder Theorem, when p(x) is divided by x + 3, the remainder is f(-1).

Remainder :

= f(-1)

= (-1)3 + 3(-1)2 + 3(-1) + 1

= -1 + 3(1) - 3 + 1

= -1 + 3 - 3 + 1

= 0

f(x) = x3 - 3x + 1

Equate the divisor (2 - 3x) to zero and solve for x.

2 - 3x = 0

x =

By Remainder Theorem, when f(x) is divided by 2 - 3x, the remainder is f().

Remainder :

= f()

= ()3 - 3() + 1

= ⁸⁄₂₇ - 2 + 1

= ⁸⁄₂₇ - 1

= ⁸⁄₂₇ - ²⁷⁄₂₇

= ⁽⁸ ⁻ ²⁷⁾⁄₂₇

= -¹⁹⁄₂₇

Let p(x) = 2x4 + 3x3 + 2kx2 + 3x + 6.

Equate the divisor (x + 2) to zero and solve for x.

x + 2 = 0

x = -2

When p(x) is divisible by (x + 2), the remainder p(-2) must be zero.

p(-2) = 0

2(-2)4 + 3(-2)3 + 2k(-2)2 + 3(-2) + 6 = 0

2(16) + 3(-8) + 2k(4) - 6 + 6 = 0

32 - 24 + 8k - 6 + 6 = 0

8 + 8k = 0

8k = -8

k = -1

Let p(x) = x3 - 4x2 - 2x + 20.

Equate (x + 2) to zero and solve for x.

x + 2 = 0

x = -2

By Factor Theorem, if (x + 2) is factor of p(x), then p(-2) must be zero.

p(-2) = (-2)3 - 4(-2)2 - 2(-2) + 20

p(-2) = -8 - 4(4) + 4 + 20

p(-2) = -8 - 16 + 4 + 20

p(-2) = 0

Therefore, (x + 2) is a factor of (x3 - 4x2 - 2x + 20).

Let f(x) = 3x3 + x2 - 20x + 12.

Equate (3x - 2) to zero and solve for x.

3x - 2 = 0

3x = 2

x =

By Factor Theorem, if (3x - 2) is factor of f(x), then f() must be zero.

f() = 3()3 + ()2 - 20() + 12

f() = 3(⁸⁄₂₇) + ⁴⁄₉ - ⁴⁰⁄₃ + 12

f() = ⁄₉ + ⁴⁄₉ - ⁴⁰⁄₃ + 12

f() = ⁄₉ + ⁴⁄₉ - ¹²⁰⁄₉ + ¹⁰⁸⁄₉

f() = ⁽⁸ ⁺ ⁴ ⁻ ¹²⁰ ⁺ ¹⁰⁸⁾⁄₉

f() = ⁽¹²⁰ ⁻ ¹²⁰⁾⁄₉

f() = 0

Therefore, (3x - 2) is a factor of (3x3 + x2 - 20x + 12)

Let f(x) = 2x3 - 6x2 + mx + 4.

Equate (x - 2) to zero and solve for x.

x - 2 = 0

x = 2

By Factor Theorem, if (x - 2) is factor of f(x), then f(2) must be equal to zero.

f(2) = 0

2(2)3 - 6(2)2 + m(2) + 4 = 0

2(8) - 6(4) + 2m + 4 = 0

16 - 24 + 2m + 4 = 0

2m - 4 = 0

2m = 4

m = 2.

p(x) = 4x3 - kx + k

Equate (x + 1) to zero and solve for x.

x + 1 = 0

x = -1

If (x + 1) divides p(x) without remainder, then the remainder is zero. That is p(-1) = 0

p(-1) = 0

4(-1)3 - k(-1) + k = 0

4(-1) + k + k = 0

-4 + 2k = 0

2k = 4

k = 2

p(x) = (3x2 - 5)(x + m) - 20

If x is a factor of p(x), then p(0) must be equal to zero.

p(0) = 0

(3(0)2 - 5)(0 + m) - 20 = 0

(0 - 5)m - 20 = 0

-5m - 20 = 0

-5m = 20

m = -4

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