Problem 1 :
Given that x − 2 is a factor of the polynomial
x^{3} − kx^{2} − 24x + 28
find k and the roots of this polynomial.
Solution :
x - 2 is a factor, then x = 2 is a zero.
Let p(x) = x^{3} − kx^{2} − 24x + 28
p(2) = 2^{3} − k(2)^{2} − 24(2) + 28
0 = 8 - 4k - 48 + 28
0 = 36 - 48 - 4k
0 = -12 - 4k
-4k = 12
k = -3
Applying the value of k, we get x^{3} + 3x^{2} − 24x + 28
The quotient = x^{2} + 5x - 14
By factoring this quotient, we will get the factors.
(x + 7) (x - 2)
So, the factors are (x + 7)(x - 2) and (x - 2).
Problem 2 :
Find the quadratic whose roots are −1 and 1/3 nd whose value at x = 2 is 10.
Solution :
Roots are x = -1 and x = 1/3
So, the factors are (x + 1) and (x - 1/3)
p(x) be the required polynomial.
p(x) = a(x + 1) (x - 1/3)
p(2) = a(2+1)(2-1/3)
10 = a(3)(5/3)
a = 10/5
a = 2
By applying the value of a in p(x), we get
So, the required quadratic polynomial is
p(x) = 2(x + 1) (x - 1/3)
Problem 3 :
Consider the polynomial p(x) = x^{3} − 4x^{2} + ax − 3.
(a) Find a if, when p(x) is divided by x + 1, the remainder is −12.
(b) Find all the factors of p(x).
Solution :
x + 1 = 0
x = -1
p(-1) = (-1)^{3} − 4(-1)^{2} + a(-1) − 3
0 = -1 - 4 - a - 3
0 = -8 - a
a = -8
By applying the value of a in the polynomial p(x), we get
p(x) = x^{3} − 4x^{2} - 8x − 3
(x + 1) and (x^{2} - 5x - 3) are factors.
Problem 4 :
Consider the polynomial
h(x) = 3x^{3} − kx^{2} − 6x + 8
(a) Given that x − 4 is a factor of h(x), find k and find the other factors of h(x).
Solution :
x - 4 is a factor, then x = 4 is a zero.
h(4) = 3(4)^{3} − k(4)^{2} − 6(4) + 8
0 = 3(64) − 16k − 24 + 8
0 = 192 - 16k - 16
0 = 176 - 16k
16k = 176
k = 11
Problem 5 :
Find the quadratic which has a remainder of −6 when divided by x − 1, a remainder of −4 when divided by x − 3 and no remainder when divided by x + 1
Solution :
Let p(x) = k (x - a) (x - b) the required quadratic polynomial.
p(x) = k (x - a) (x - b) ----(1)
When p(x) is divided by x - 1, the remainder will be -6
x - 1 = 0
x = 1 an p(1) = -6
Applying the value x = 1 in (1), we get
p(1) = k(1 - a)(1 - b)
-6 = k(1 - a) (1 - b) ---(2)
When p(x) is divided by x - 3, the remainder will be -4
x - 3 = 0
x = 3 an p(3) = -4
Applying the value x = 3 in (1), we get
p(3) = k(3 - a)(3 - b)
-4 = k(3 - a) (3 - b) ---(3)
No remainder, that is p(-1) = 0
p(-1) = k(-1 - a)(-1 - b)
0 = k(-1 - a)(-1 - b)
(-1-a) (-1-b) = 0
-1-a = 0
a = -1
Applying the value of a in (2), we get
-6 = k(1+1) (1 - b)
-3 = k(1 - b)
k = -3/(1 - b)
Applying the value of a in (3), we get
-4 = k(3 + 1) (3 - b)
-1 = k(3 - b)
k = -1/(3 - b)
Equating k terms,
-3/(1 - b) = -1/(3 - b)
-3(3 - b) = -1(1 - b)
-9 + 3b = -1 + b
2b = 8
b = 4
k = -1/(3 - 4)
k = 1
p(x) = 1 (x - (-1)) (x - 4)
p(x) = 1 (x + 1) (x - 4)
p(x) = x^{2} - 3x - 4
Problem 6 :
Find the value of a if x − 3 is a factor of f(x) = x^{3} - 11x + a
Solution :
x - 3 is a factor, then x = 3 is the zero.
f(3) = 3^{3} - 11(3) + a
0 = 27 - 33 + a
0 = -6 + a
a = 6
Problem 7 :
Find the value of k if f(x) = 3(x^{2} + 3x - 4) - 8(x - k) is divisible by x.
Solution :
It is divisible by x, then x - 0 is a factor.
x = 0
f(0) = 3(0^{2} + 3(0) - 4) - 8(0 - k)
0 = 3(-4) - 8(-k)
0 = -12 + 8k
8k = 12
k = 12/8
k = 3/2
Problem 8 :
If x − 2 is a factor of polynomial
p(x) = a(x^{3} - 2x) + b(x^{2} - 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 :
Since x - 2 is a factor, then x = 2 is a zero of the polynomial.
p(2) = a(2^{3} - 2(2)) + b(2^{2} - 5)
0 = a(8 - 4) + b(4 - 5)
0 = 4a - b
4a - b = 0
So, option d is correct.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM