Problem 1 :
The expression
is equal to which of the following ?
Solution :
We cannot divide the polynomials using long division, because it is not an improper fraction.
Simplifying option b :
Simplifying option c :
So, option c is correct.
Problem 2 :
If the expression
is written in the form
what us Q in terms of x ?
(a) 3x - 1 (b) 3x + 1 (c) 6x^{2} + 3x + 1 (c) 6x^{2} + 5x + 1
Solution :
So, the value of Q is 3x + 1.
Problem 3 :
The expression 4x^{2} + 5 can be written as A(2x - 1) + R, where A is an expression in terms of x and R is constant. What is the value of R?
Solution :
Using long division,
Using division algorithm,
Dividend = divisor x quotient + remainder
= (2x - 1)(2x + 1) + 6
The value of R is 6.
Problem 4 :
The function g is defined by a polynomial. The table above shows some values of x and g(x). What is the remainder when g(x) is divided by x + 3 ?
(a) -2 (b) 1 (c) 2 (d) 6
Solution :
When x + 3 =0
x = -3
When the polynomial divided by x + 3, we get the remainder 2.
Problem 5 :
2z^{3} - kxz^{2} + 5xz + 2x - 2
In the polynomial above, k is constant. If z - 1 is a factor of the polynomial above, what is the value of k ?
Solution :
z - 1 = 0
z = 1
Let p(z) = 2z^{3} - kxz^{2} + 5xz + 2x - 2
p(1) = 2(1)^{3} - kx(1)^{2} + 5x(1) + 2x - 2
p(1) = 2 - kx + 5x + 2x - 2
0 = -kx + 7x
k = 7
Problem 6 :
When 3x^{2} - 8x - 4 is divided by 3x - 2, the result can be expresses as
What is A in terms of x ?
(a) x - 4 (b) x - 2 (c) x + 2 (d) x + 4
Solution :
So, the value of A is x - 2.
Problem 7 :
The expression 2x^{2} - 4x - 3 can be written as A(x + 1) + B, where B is constant. What is A in terms of x ?
(a) 2x + 6 (b) 2x + 2 (c) 2x - 2 (d) 2x - 6
Solution :
From the question, it is clear that 2x^{2} - 4x - 3 should be divided by x + 1
So, the value of A is 2x - 6.
Problem 8 :
The expression x^{2} + 4x - 9 can be written as (ax + b)(x - 2) + c, where a, b and c are constants. What is the value of a + b + c ?
(a) -2 (b) 3 (c) 7 (d) 10
Solution :
ax + b = x + 6 and c = 3
a = 1, b = 6 and c = 3
a + b + c = 1 + 6 + 3 ==> 10
So, the answer is 10.
Problem 9 :
For a polynomial p(x), p(2) = 0. Which of the following must be true about p(x) ?
(a) 2x is a factor of p(x) (b) 2x - 2 is a factor of p(x)
(c) x - 2 is a factor of p(x) (d) x + 2 is a factor of p(x)
Solution :
p(2) = 0
x = 2
x - 2
Then x - 2 is a factor.
Problem 10 :
If p(x) = x^{3} + x^{2} - 5x + 3, then p(x) is divisible by which of he following ?
I. x - 2 II. x - 1 III. x + 3
(a) I and II only (b) I and III only
(c) II and III only (d) I, II and III
Solution :
I. x - 1
x - 1 = 0
x = 1
Let us check if x - 1 is a factor of the polynomial, p(x).
x - 1 is a factor.
II. x - 2
x - 2 = 0
x = 2
Let us check if x - 2 is a factor of the polynomial, p(x).
x - 2 is not a factor.
III. x + 3
x + 3 = 0
x = -3
Let us check if x + 3 is a factor of the polynomial, p(x).
x + 3 is a factor.
So, I and III are factors.
Problem 11 :
If the polynomial p(x) is divisible by x - 2, which of the following could be p(x)
(a) p(x) = -x^{2} + 5x - 14 (b) p(x) = x^{2} - 6x - 2
(c) p(x) = 2x^{2} + x - 8 (d) p(x) = 3x^{2} - 2x - 8
Solution :
Option (a) :
p(x) = -x^{2} + 5x - 14
p(x) = -(x^{2} - 5x + 14)
p(x) = -(x - 7)(x + 2)
Option (b) :
p(x) = x^{2} - 6x - 2
Since it is not factorable, x - 2 is not a factor.
Option (d) :
p(x) = 3x^{2} - 2x - 8
p(x) = (3x^{2} - 6x + 4x - 8)
p(x) = 3x(x - 2) + 4(x - 2)
p(x) = (3x + 4)(x - 2)
So, the factor of 3x^{2} - 2x - 8 is x - 2.
Problem 12 :
If x - 1 and x + 1 are both factors of the polynomial
ax^{4} + bx^{3} - 3x^{2} + 5x
and a and b are constants, what is the value of a ?
(a) -3 (b) 1 (c) 3 (d) 5
Solution :
Let p(x) = ax^{4} + bx^{3} - 3x^{2} + 5x
x - 1 is a factor, then x = 1 is a solution.
p(1) = a(1)^{4} + b(1)^{3} - 3(1)^{2} + 5(1)
0 = a + b - 3 + 5
a + b = -2 -----(1)
x + 1 is a factor, then x = -1 is a solution.
p(-1) = a(-1)^{4} + b(-1)^{3} - 3(-1)^{2} + 5(-1)
0 = a - b - 3 - 5
a + b = 8 -----(2)
(1) + (2)
2a = 6
a = 3
So, the answer is 3.
Problem 13 :
For a polynomial p(x), p(1/3) = 0. Which of the following must be a factor of p(x)
(a) 3x - 1 (b) 3x + 1 (c) x - 3 (d) x + 3
Solution :
Since 1/3 is a zero of the polynomial.
x = 1/3 is a solution.
Converting it as factor, we get x - 1/3
(3x - 1) is a factor of p(x)
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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