Determine if d(x) is a factor of f(x).
Problem 1 :
f(x) = x^{3} + 3x^{2} - 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) = 3x^{4} + 7x^{3} + 3x^{2} - 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) = 2x^{3} – x^{2} - 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) = 4x^{3} - 2x^{2} + 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) = 3x^{4} - 3x^{3} - 9x^{2} + 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) = ax^{2} – 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) = ax^{2} – 3(a – 1)x – 1
p(1) = ax^{2} – 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) = 6x^{3} + 5x^{2} - 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 x^{3} – 2mx^{2} + 16 divisible by x + 2
Solution :
Let p(x) = x^{3} – 2mx^{2} + 16
P(x) is divisible by x + 2.
x + 2 = 0
x = -2
p(-2) = x^{3} – 2mx^{2} + 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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