Zeros and Coefficients of Polynomial - Class 10 Maths MCQ

This set of Class 10 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Zeros and Coefficients of Polynomial – 1”.

1. The zeros of the polynomial 18x²-27x+7 are ___________
a) \(\frac {7}{6}, \frac {1}{3}\)
b) \(\frac {-7}{6}, \frac {1}{3}\)
c) \(\frac {7}{6}, \frac {-1}{3}\)
d) \(\frac {7}{3}, \frac {1}{3}\)
View Answer

Answer: a
Explanation: 18x²-27x+7=0
18x²-21x-6x+7=0
3x(6x-7)-1(6x-7)=0
(6x-7)(3x-1)=0
x=\(\frac {7}{6}, \frac {1}{3}\)
The zeros are \(\frac {7}{6}\) and \(\frac {1}{3}\).

2. What will be the polynomial if its zeros are 3, -3, 9 and -9?
a) x⁴-80x²+729
b) x⁴-90x²+729
c) x⁴-90x²+79
d) x⁴-100x²+729
View Answer

Answer: b
Explanation: The zeros of the polynomial are 3, -3, 9 and -9.
Then, (x-3), (x+3), (x-9) and (x+9) are the factors of the polynomial.
Multiplying the factors, we have
(x-3) (x+3) (x-9) (x+9)
(x²-9) (x²-81) (By identity (x-a)(x+a)=x²-a²)
(x⁴-9x²-81x²+729)
x⁴-90x²+729

3. The sum and product of zeros of a quadratic polynomial are 10 and \(\frac {5}{2}\) respectively. What will be the quadratic polynomial?
a) 2x²-20x+10
b) 2x²-x+5
c) 2x²-20x+5
d) x²-20x+5
View Answer

Answer: c
Explanation: The sum of the polynomial is 10, that is, α+β = 10
The product of the polynomial is \(\frac {5}{2}\) i.e. αβ = \(\frac {5}{2}\)
∴ f(x)=x²-(α+β)x+αβ
f(x)=x²-10x+\(\frac {5}{2}\)
f(x)=2x²-20x+5

4. If α and β are the zeros of x²+20x-80, then the value of α+β is _______
a) -15
b) -5
c) -10
d) -20
View Answer

Answer: d
Explanation: α and β are the zeros of x²+20x-80.
Sum of zeros or α+β = \(\frac {-coefficient \, of \, x}{coefficient \, of \, x^2} = \frac {-20}{1}\) = -20

5. If α and β are the zeros of 3x²-5x-15, then the value of αβ is _______
a) -5
b) -10
c) -15
d) -20
View Answer

Answer: a
Explanation: α and β are the zeros of 3x²-5x-15.
Product of zeros or αβ = \(\frac {constant \, term}{coefficient \, of \, x^2} = \frac {-15}{3}\) = -5

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6. What will be the value of other zero, if one zero of the quadratic polynomial is 5 and the sum of the zeros is 10?
a) 10
b) 5
c) -5
d) -10
View Answer

Answer: b
Explanation: One zero of the quadratic polynomial is 5. ∴ the factor of the polynomial is (x-5)
Let us assume the other zero to be b. ∴ the other factor of the polynomial is (x-b)
Multiplying the factors, we have (x-5)(x-b)
x²-5x-bx+5b
x²-(5+b)x+5b
The sum of zeros is 10.
∴ \(\frac {-coefficient \, of \, x}{coefficient \, of \, x^2}\)=10
\(\frac {-(-5-b)}{1}\)=10
5+b=10
b=5
The equation becomes x²-10x+25.
Therefore, the other zero is 5.

7. The value of a and b, if the zeros of x²+(a+5)x-(b-4) are -5 and 9 will be _________
a) 47, -5
b) -5, 47
c) -9, 49
d) -4, 45
View Answer

Answer: c
Explanation: The zeros of the polynomial are -5 and 9.
Hence, α=-5, β=9
The polynomial is x²+(a+5)x-(b-4).
Sum of zeros or α+β=-5+9 = \(\frac {-coefficient \, of \, x}{coefficient \, of \, x^2} = \frac {a+5}{1}\)
-4=a+5
a = -9
Product of zeros or αβ = -45 = \(\frac {constant \, term}{coefficient \, of \, x^2} = \frac {-(b-4)}{1}\)
-45=-b+4
b=49

8. What will be the value of k, if one zero of x²+(k-3)x-16=0 is additive inverse of other?
a) 4
b) -4
c) -3
d) 3
View Answer

Answer: d
Explanation: Since, one zero of the polynomial is the additive inverse of the other.
Hence, the sum of roots will be zero.
The polynomial is x²+(k-3)x-16=0
Sum of zeros or α+β=\(\frac {-coefficient \, of \, x}{coefficient \, of \, x^2} = \frac {k-3}{1}\)=0
k-3=0
k=3

9. If α and β are the zeros of 10x²+20x-80, then the value of \(\frac {1}{\alpha } + \frac {1}{\beta }\) is _______
a) \(\frac {5}{4}\)
b) \(\frac {1}{5}\)
c) \(\frac {3}{4}\)
d) \(\frac {1}{4}\)
View Answer

Answer: d
Explanation: \(\frac {1}{\alpha } + \frac {1}{\beta } = \frac {\alpha +\beta }{\alpha \beta }\)
α+β=\(\frac {-20}{10}\)=-2
αβ=\(\frac {-80}{10}\)=-8
∴ \(\frac {\alpha +\beta }{\alpha \beta } = \frac {-2}{-8} = \frac {1}{4}\)

10. If α and β are the zeros of x²+35x-75, then _______
a) α+β<αβ
b) α+β>αβ
c) α+β=αβ
d) α+β≠αβ
View Answer

Answer: b
Explanation: The given polynomial is x²+35x-75.
The sum of zeros, α + β = \(\frac {-coefficient \, of \, x}{coefficient \, of \, x^2} = \frac {-35}{1}\) = -35
The product of zeros, αβ = \(\frac {constant \, term}{coefficient \, of \, x^2}\) = -75
Clearly, sum of zeros is greater than product of zeros.

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