Free Polynomials 02 Practice Test - 9th Grade
Question 1
If xg+1xa is a polynomial, what could be the value of a?
only 1
any positive integer
any negative integer
any integer
SOLUTION
Solution : C
For an expression to be a polynomial, the exponents of x should be whole numbers. The above expression can be written as xg+x−a . For this to be a polynomial, −a should be a whole number. This is possible if a is a negative integer or zero.
Question 2
If ax2+bx+c is a monomial in x , what can be said about a, b and c?
Atleast one of a, b and c is non - zero.
Only one among a, b and c is zero.
Only one among a, b and c is non - zero.
All three should be non - zero.
SOLUTION
Solution : C
Since a monomial can contain only one term, only one of a, b and c can be non-zero to leave a single term.
Question 3
If axn is a zero degree polynomial in x, what can be said about a?
a is negative integer always
a is 1 always
a is 0 always
a is non zero
SOLUTION
Solution : D
A zero degree polynomial is essentially a constant. i.e., any non-zero real number.
So, if axn is a zero degree polynomial, then
axn = ax0 = a
So, n has to be zero, and a should not be zero.
Question 4
Find the factors of x3+2x2+2x+1
SOLUTION
Solution : A and B
Let p(x)=(x3+2x2+2x+1) be the given polynomial
By observation, we can see that by putting
x=−1, p(x)=0
∴x=−1 is the root of p(x)
⇒(x+1) is a factor of p(x)
To find the other factor, we perform long division of p(x) by (x+1)
∴ The two factors of p(x) are (x+1) and (x2+x+1)
Question 5
Given the area of rectangle is A=25a2−35a+69. The length is given as (5a−3).Find the width of the rectangle.
5a−3
5a−4
4a−5
a−4
SOLUTION
Solution : B
Let y be the width.
Given: Area =25a2−35a+69
We know that, area of a rectangle = Length ×breadth
Hence, y×(5a−3)=25a2−35a+69
y=(25a2−35a+69)÷(5a−3)
By long division method,
5a−35a−23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a2−35a+69 25a2−12a − + ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ −23a+69 −23a+69 + − ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0
∴y=5a−23Width of the rectangle =5a−23
Question 6
If a polynomial(x2+ax−c)cuts x-axis at 2 and -4, the value of a+c is
SOLUTION
Solution :Suppose f(x) is the polynomial.
Now since it cuts x-axis at 2 and -4, the roots of f(x) are 2 and -4.
Sum of roots = −a=2−4=−2
⇒a=2
Product of roots = −c=2×−4=−8
⇒c=8
Hence, (a+c)=10
Question 7
Which of the following are the factors of 1012−992
SOLUTION
Solution : B and C
Using the algebraic Identity (a2−b2)=(a−b)(a+b)
⇒(a−b) and (a+b) are factors of a2−b2.
Consider,
(1012−992)=(101+99)(101−99)
=(200)(2)
We find that 2 and 200 both are the factors of (1012−992).
Question 8
Factor Theorem is equivalent to Remainder Theorem when remainder is
SOLUTION
Solution : C
If f(x) is a polynomial and is divided by (x-a), then if f(a) = k, this k will be the remainder, as stated by the Remainder Theorem. If the value of k is 0, then (x-a) is the factor of f(x), as indicated by the Factor Theorem. Hence, factor theorem is equivalent to remainder theorem when remainder is 0.
Question 9
Select the values which the coefficients and exponents of a polynomial can take? (Options are given by comma separating them in the form - (coefficient,exponent))
1.68686868... , 0
SOLUTION
Solution : A and C
Any real number can be a coefficient in a polynomial. Thus, rational and irrational numbers are valid coefficients.
Exponents in polynomials can only be whole numbers.
Question 10
If f(x) = (x - a)(x - b) , then if a and b are the zeros of the polynomial f(x), f(a) = f(b).
True
False
SOLUTION
Solution : A
Since a and b are the zeroes of the polynomial f(x), both f(a) = 0 and f(b) = 0, i.e. a and b will be the roots of the equation f(x) = 0.
Thus, f(a) = f(b) = 0. Hence, the given statement is true.