# CHECK IF THE LINEAR IS A FACTOR OF CUBIC POLYNOMIAL

Determine if d(x) is a factor of f(x).

Problem 1 :

f(x) = x3 + 3x2 - 6x - 18

d(x) = x + 3

Solution :

To check if d(x) is a factor of the given polynomial f(x).

x + 3 = 0

x = -3

f(-3) = (-3)3 + 3(-3)2 - 6(-3) - 18

f(-3) = -27 + 27 + 18 - 18

f(-3) = 0

Since the remainder is 0, d(x) is the factor of the polynomial f(x).

Determine if d(x) is a factor of f(x).

Problem 2 :

f(x) = 3x4 + 7x3 + 3x2 - x – 4

d(x) = x + 2

Solution :

x + 2 = 0

x = -2

f(-2) = 3(-2)4 + 7(-2)3 + 3(-2)2 – (-2) – 4

= 48 - 56 + 12 + 2 - 4

= 2

Since the remainder is not equal to zero, we can decide x + 2 is not a factor of the given polynomial f(x).

Problem 3 :

f(x) = 2x3 – x2 - 6x - 1

d(x) = x + 1

Solution :

x + 1 = 0

x = -1

f(-1) = 2(-1)3 – (-1)2 – 6(-1) - 1

= -2 - 1 + 6 - 1

= 2

Since the remainder is not equal to zero, we can decide x + 1 is not a factor of the given polynomial f(x).

Problem 4 :

f(x) = 4x3 - 2x2 + x - 3

d(x) = x - 1

Solution :

x - 1 = 0

x = 1

f(1) = 4(1)3 – 2(1)2 + 1 - 3

= 4 - 2 + 1 - 3

= 0

Since the remainder is equal to zero, we can decide x - 1 is a factor of the given polynomial f(x).

Problem 5 :

f(x) = 3x4 - 3x3 - 9x2 + 5x – 2

d(x) = x - 2

Solution :

x - 2 = 0

x = 2

f(2) = 3(2)4 – 3(2)3 – 9(2)2 + 5(2) – 2

= 48 - 24 - 36 + 10 - 2

= -4

Since the remainder is not equal to zero, we can decide x – 2 is not a factor of the given polynomial f(x).

Problem 6 :

If x – 1 is a factor of the polynomial

p(x) = ax2 – 3(a – 1)x – 1, then find the value of ‘a’?

Solution :

Given, x – 1 is a factor of the polynomial

x – 1 = 0

x = 1

p(x) = ax2 – 3(a – 1)x – 1

p(1)  = ax2 – 3ax + 3x – 1

= a(1)2 – 3a(1) + 3(1) – 1

= a – 3a + 3 – 1

= -2a + 2

-2a = -2

Dividing -2 on both sides.

-2/-2a = -2/-2

a = 1

Hence, the value of a is 1.

Problem 7 :

Check If x + 1 is a factor of polynomial, hence evaluate p(-1).

p(x) = 6x3 + 5x2 - 3x + 2,

Solution :

Given, x + 1 is a factor of polynomial

x + 1 = 0

x = -1

p(-1) = 6(-1)3 + 5(-1)2 – 3(-1) + 2,

= -6 + 5 + 3 + 2

p (-1) = 4

Hence, the value of p(-1) is 4.

Problem 8 :

For what value of m is x3 – 2mx2 + 16 divisible by x + 2

Solution :

Let p(x) = x3 – 2mx2 + 16

P(x) is divisible by x + 2.

x + 2 = 0

x = -2

p(-2) = x3 – 2mx2 + 16

= (-2)3 – 2m(-2)2 + 16

= -8 – 2m(4) + 16

= -8 – 8m + 16

= 8 – 8m

8m = 8

Dividing 8 on both sides.

8m/8 = 8/8

m = 1

So, the value of m is 1.

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