Free Polynomials 02 Practice Test - 10th Grade
Question 1
Zeroes of x2−56x+108 are______.
both positive
both negative
one negative and one positive
both equal
SOLUTION
Solution : A
In a quadratic equation ax2+bx+c,both zeroes are positive when a is positive, b is negative and c is positive.
In this case, x2−56x+108 can be factorized as (x−2)(x−54).
⟹the roots are 2 and 54; both are positive.
Question 2
If α,β and γ are the zeroes of the polynomial f(x)=ax3+bx2+cx+d, then 1α+1β+1γ is _____.
cd
ad
−cd
−ba
SOLUTION
Solution : C
Given α,β and γ are the zeroes of the polynomial f(x)=ax3+bx2+cx+d
1α+1β+1γ
=βγ+αγ+αβ(αβγ) =(ca)(−da) =−cd
Question 3
When 3x3+2x2−x+4 is divided by 3+2x the remainder is −18.
True
False
SOLUTION
Solution : A
P(x)=3x3+2x2−x+4
If P(x) is divided by (x−a), then the remainder is P(a)
2x+3=0⇒x=−32
So, the remainder is the value of polynomial at −32
Substituting x=−32 in p(x)
p(−32)=3(−32)3+2(−32)2−(−32)+4
p(−32)=3(−278)+2(94)−(−32)+4
p(−32)=−818+92+32+4
p(−32)=−81+36+12+328
p(−32)=−81+808
p(−32)=−18
∴ The remainder is −18.
Thus, the given statement is true.
Question 4
What will be the remainder when 4x4−3x3+2x2−x is divided by x+1?
1
0
7
10
SOLUTION
Solution : D
Given p(x)=4x4−3x3+2x2−x
To find the remainder of 4x4−3x3+2x2−xx+1,
we will use remainder theorem.
Remainder of p(x)x−a is p(a).
⇒ Remainder of 4x4−3x3+2x2−xx+1 is p(−1).[∵x+1=x−(−1)]
p(−1)=4(−1)4−3(−1)3+2(−1)2−(−1)
⇒p(−1)=4+3+2+1=10Hence, the remainder of 4x4−3x3+2x2−xx+1 is 10.
Question 5
Which of the following graph represents a quadratic polynomial which has sum of its zeroes is zero?
A)
B)
C)
D)
SOLUTION
Solution : D
We know that, the values at which a quadratic polynomial cuts the x-axis are its zeroes.
Option A: Polynomial cuts the x-axis at : -3, -2 ; sum of zeroes: -5
Option B: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option C: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option D: Polynomial cuts the x-axis at : -2, 2; sum of zeros: 0
Hence, the answer is D.
Question 6
If 1 is a zero of the polynomial p(x), then which of the following is correct?
p(1) = x
p(1) = 0
p(1) = 1
p(2) - p(1) = p(2)
SOLUTION
Solution : B and D
If k is a zero of the polynomial p(x), then, p(k)=0.
Here, 1 is a zero of the polynomial p(x).
⇒p(1)=0
Also, p(2)−p(1)=p(2)−0=p(2)
Question 7
The zeroes of mn(x2+1)=(m2+n2)x are:
SOLUTION
Solution : A
Given polynomial, mn(x2+1)=(m2+n2)x
Let's factorise it.
mnx2−(m2+n2)x+mn=0
⇒mnx2−m2x−n2x+mn=0
⇒mx(nx−m)−n(nx−m)=0
⇒(mx−n)(nx−m)=0
⇒mx−n=0 or nx−m=0
⇒x=mn or nm
So, zeroes of the given polynomial are mn and nm.
Question 8
If α,β are the zeroes of the polynomial x2−px+36 and α2+β2 = 9, then what is the value of p?
±6
±3
±8
±9
SOLUTION
Solution : D
Given polynomial x2−px+36
On comparing with the standard form of a quadratic polynomial ax2+bx+c, we get
a = 1, b = -p, c = 36
Here, α and β are the zeroes of the polynomial.
⇒α+β=−ba=p
and αβ=ca=36
Now, α2+β2=(α+β)2 - 2αβ
⇒ 9=p2−2×36 [∵α2+β2 = 9]
⇒81=p2
⇒p=9 or −9
Question 9
Which of the following graph represents the quadratic polynomial −x2+5x−6?
SOLUTION
Solution : C
One way of solving this question is to find the zeroes of given polynomial by conventional factorization. But we will solve it in a smart way. We just find the sum of zeroes from polynomial and compare it from the graphs, which satisfies the sum of zeroes.
From polynomial,
Sum of zeroes = −ba = −5−1 = 5
From the graphs given , we look for graph whose sum of zeroes is 5. Thus graph having 2,3 as zeroes is the required graph.
Question 10
Graph of a quadratic polynomial is a
SOLUTION
Solution : C
Quadratic polynomial is of degree 2 and the graph of a quadratic polynomial is parabola.