Problem 1 :
If y = 1 is a zero of the polynomial
q(y) = 4y^{3} + ky^{2} - y - 1
then find the value of k.
Solution :
Since y = 1 is one of the zero, it will satisfy the polynomial q(y).
q(y) = 4y^{3} + ky^{2} - y - 1
q(1) = 4(1)^{3} + k(1)^{2} - 1 - 1
0 = 4 + k - 2
0 = 2 + k
k = -2
So, the value of k is -2.
Problem 2 :
For what value of m is x^{3} - 2mx^{2} + 16 is divisible by x + 2.
Solution :
Since the given polynomial x^{3} - 2mx^{2} + 16 is divisible by x + 2, then p(-2) = 0
p(x) = x^{3} - 2mx^{2} + 16
p(-2) = (-2)^{3} - 2m(-2)^{2} + 16
0 = -8 - 8m + 16
0 = 8 - 8m
8m = 8
m = 8/8
m = 1
Problem 3 :
The polynomials x^{3} + 2x^{2} - 5ax - 7 and x^{3} + ax^{2} - 12x + 6 when divided by x + 1 and x - 2 respectively, leave remainders R_{1} and R_{2} respectively. Find the value of a in each of the following cases :
i) R_{1} = R_{2}
ii) R_{1} + R_{2} = 0
iii) 2R_{1} + R_{2} = 0
Solution :
Let p(x) = x^{3} + 2x^{2} - 5ax - 7 and q(x) = x^{3} + ax^{2} - 12x + 6
x + 1 = 0, then x = -1
Finding remainder p(x) = x^{3} + 2x^{2} - 5ax - 7 by x + 1
p(-1) = (-1)^{3} + 2(-1)^{2} - 5a(-1) - 7
R_{1} = -1 + 2 + 5a - 7
R_{1} = -8 + 2 + 5a
R_{1} = -6 + 5a
x - 2 = 0, then x = 2
Finding remainder q(x) = x^{3} + ax^{2} - 12x + 6 by x + 2
q(2) = 2^{3} + a(2)^{2} - 12(2) + 6
R_{2} = 8 + 4a - 24 + 6
R_{2} = 14 - 24 + 4a
R_{2} = - 10 + 4a
i) When R_{1} = R_{2}
-6 + 5a = -10 + 4a
5a - 4a = -10 + 6
a = -4
ii) R_{1} + R_{2} = 0
-6 + 5a - 10 + 4a = 0
9a - 16 = 0
a = 16/9
iii) 2R_{1} + R_{2} = 0
2(-6+ 5a) + (-10 + 4a) = 0
-12 + 10a - 10 + 4a = 0
14a - 22 = 0
a = 22/14
a = 11/7
Problem 4 :
When a polynomial
p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + b
is divisible by x - 1 and x + 1, the remainders are 5 and 19 respectively. Find the remainder when p(x) is divided by x - 2.
Solution :
p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + b
When the polynomial p(x) is divided by x - 1, the remainder is 5.
x - 1 = 0
x = 1
p(1) = 1^{4} - 2(1)^{3} + 3(1)^{2} - a(1) + b
5 = 1 - 2 + 3 - a + b
5 = 2 - a + b
-a + b = 5 - 2
-a + b = 3 -----(1)
When the polynomial p(x) is divided by x + 1, the remainder is 19.
x + 1 = 0
x = -1
p(-1) = (-1)^{4} - 2(-1)^{3} + 3(-1)^{2} - a(-1) + b
19 = 1 + 2 + 3 + a + b
19 = 6 + a + b
a + b = 19 - 6
a + b = 13 ----(2)
(1) + (2)
2b = 3 + 13
2b = 16
b = 8
Applying the value of b in (1), we get
-a + 8 = 3
-a = 3 - 8
-a = -5
a = 5
So, the values of a and b are 5 and 8 respectively.
Applying the values of a and b, we get
p(x) = x^{4} - 2x^{3} + 3x^{2} - 5x + 8
Dividing p(x) by x - 2, we get
x - 2 = 0
x = 2
p(2) = 2^{4} - 2(2)^{3} + 3(2)^{2} - 5(2) + 8
p(2) = 16 - 16 + 12 - 10 + 8
p(2) = 10
Problem 5 :
If x - 3 and x - 1/3 are both factors of ax^{2} + 5x + b, show that a = b.
Solution :
Let p(x) = ax^{2} + 5x + b
When x - 3 = 0, then x = 3
p(3) = a(3)^{2} + 5(3) + b
p(3) = 9a +15 + b
0 = 9a +15 + b
9a + b = -15 ------(1)
When x - 1/3 = 0, then x = 1/3
p(1/3) = a(1/3)^{2} + 5(1/3) + b
p(1/3) = a/9 +5/3 + b
0 = a/9 +5/3 + b
a/9 + b = -5/3 ------(2)
(1) - (2)
9a - a/9 = -15 + (5/3)
a = -3/2
By applying the value of a in (1), we get
9(-3/2) + b = -15
-27/2 + b = -15
b = -15 - (-27/2)
b = -3/2
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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