This set of Class 10 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Division of Polynomial”.
1. If f(x) is divided by g(x), it gives quotient as q(x) and remainder as r(x). Then, f(x)=q(x)×g(x)+r(x) where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
a) True
b) False
View Answer
Explanation: Consider, f(x) is 27x2-39x, q(x) as 9x+2, g(x) as 3x-5 and remainder is 10.
f(x)=q(x)×g(x)+r(x)
RHS
q(x)×g(x)+r(x)=(9x+2)(3x-5)+10=27x2-45x+6x-10+10=27x2-39x, which is equal to LHS.
Hence proved.
2. If α is a zero of the polynomial f(x), then the divisor of f(x) will be _________
a) x<α
b) x-α
c) x>α
d) x+α
View Answer
Explanation: If α is a zero of the polynomial f(x).
The divisor will be x-α.
For example, if 5 is a zero of a polynomial f(x), then its divisor will be x-5.
3. If two of the zeros of the polynomial f(x)=x3+(6-√3)x2+(-1-√3)x+30-6√3 are 3 and -2 then, the other zero will be ____________
a) -√3
b) 5
c) 5-√3
d) 5+√3
View Answer
Explanation: Since the zeros of the polynomial are 3 and -2.
The divisor of the polynomial will be (x-3) and (x+2).
Multiplying (x-3) and (x+2) = x2+2x-3x-6=x2-x+6
Dividing, x3+(6-√3)x2+(-1-√3)x+30-6√3 by x2-x+6
We get, x-5+√3 as quotient.
Hence, the third zero will be 5-√3.
4. What will be the value of a and b if the polynomial f(x)=30x4-50x3+109x2-23x+25, when divided by 3x2-5x+10, gives 10x2+3 as quotient and ax+b as remainder?
a) a=8, b=5
b) a=-8, b=5
c) a=8, b=-5
d) a=-8, b=-5
View Answer
Explanation: We know that,
f(x)=q(x)×g(x)+r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
∴ 30x4-50x3+109x2-23x+25=(10x2+3)(3x2-5x+10)+ax+b
30x4-50x3+109x2-23x+25=30x4-50x3+109x2-15x+30+ax+b
30x4-50x3+109x2-23x+25-(30x4-50x3+109x2-15x+30)=ax+b
-23x+25+15x-30=ax+b
-8x-5=ax+b
∴ a=-8, b=-5
5. The quotient if the polynomial f(x)=50x2-90x-25 leaves a remainder of -5, when divided by 5x-10, will be __________
a) 10x+2
b) 10x-2
c) -10x+2
d) -10x-2
View Answer
Explanation: We know that,
f(x)=q(x)×g(x)+r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
∴ 50x2-90x-25=q(x)×5x-10-5
50x2-90x-25+5=q(x)×5x-10
\(\frac {50x^2-90x-20}{5x-10}\)=q(x)
We get, q(x)=10x+2
6. The polynomial (x), if the divisor is 5x2, quotient is 2x+3, and remainder is 10x+20 is __________
a) 10x3-15x2-10x-20
b) -10x3-15x2+10x+20
c) 10x3+15x2+10x+20
d) -10x3+15x2+10x+20
View Answer
Explanation: We know that,
f(x)=q(x)×g(x)+r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
f(x)=5x2×(2x+3)+10x+20
f(x)=10x3+15x2+10x+20
7. When a polynomial f(x)=acx3+bcx+d, is divided by g(x), it leaves quotient as cx, and remainder as d. The value of g(x)will be _____
a) -ax2+b
b) ax2-b
c) ax2+b
d) x2+b
View Answer
Explanation: We know that,
f(x)=q(x)×g(x)+r(x)
Where, f(x) is the dividend, q(x) is the quotient, g(x) is the divisor and r(x) is the remainder.
acx3 + bcx + d = cx × g(x) + d
acx3 + bcx + d – d = cx × g(x)
\(\frac {acx^3+bcx}{cx}\)=g(x)
g(x)=ax2+b
8. The real number that should be subtracted from the polynomial f(x)=15x5+70x4+35x3-135x2-40x-11 so that it is exactly divisible by 5x4+10x3-15x2-5x is ____________
a) -12
b) -11
c) 11
d) 12
View Answer
Explanation: On dividing, 15x5+70x4+35x3-135x2-40x-11 by 5x4+10x3-15x2-5x
We get, 3x+8 as quotient and remainder as -11.
So if we subtract -11 from 15x5+70x4+35x3-135x2-40x-11 it will be exactly divisible by 5x4+10x3-15x2-5x.
9. What real number that should be added to the polynomial f(x)=81x2-31, so that it is exactly divisible by 9x+1?
a) 40
b) 10
c) 30
d) 20
View Answer
Explanation: 81x2-31 is exactly divisible by 9x+1
Hence, on dividing 81x2-31 by 9x+1
We get, 9x-1 as quotient and remainder as -30.
So if we add 30 to 81x2-31, it will be exactly divisible by 9x+1.
10. If the polynomial f(x)=x2+kx-15, is exactly divisible by x-5, then the value of k is _______
a) 3
b) 2
c) -3
d) -2
View Answer
Explanation: x2+kx-15 is exactly divisible by x-5
Dividing, x2+kx+15 by x-5
We get, 5k+10 as remainder.
Since, x2+kx-15 is exactly divisible by 2x-5
∴ 5k+10=0
k=-2
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