Q 1 – If the roots of the quadratic polynomial are equal, where the discriminant D = b2 – 4ac, then
(a) D > 0
(b) D < 0
(c) D ≥ 0
(d) D = 0
(d) D = 0
Q 2 – If the sum of the roots is –p and product of the roots is –1/p, then the quadratic polynomial is
a) k(–px2 + x/p + 1)
b) k(px2 – x/p – 1)
c) k(x2 + px – 1/p)
d) k(x2 – px + 1/p)
d) k(x2 – px + 1/p)
Q 3 – If one zero of the polynomial x2– 4x +1 is 2 + √3, find the other zero.
Given : (2+ √3) is one of the polynomial x2−4x+1 to find : other zero.
since given polynomial is quadratic so it has only two zeros.
Let other zero be x.
Now, Sum of zero = 
x+2+ √3 =4
or x = 2 – √3
hence , the other zero is (2 –√3 )
Q 4 – If α and β are the zeroes of the polynomial f(x)= x2 +x +1, then 1/α + 1/β is
By taking L.C.M of dinominators
→ 1/α + 1/β = (α + β)/αβ
Since we know that,
→ α + β = – Coefficient of x/Coefficient of x²
→ α + β = – 1/1 = – 1
→ αβ = Constant Term/Coefficient of x²
→ αβ = 1/1 = 1
∴ (α + β)/αβ = – 1/1
→ – 1
Q 5 – If the polynomial f(x) = x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
The given polynomial is p(x) = x4 – 6x3 + 16x2 – 25x + 10
We can solve this by using division algorithm i.e, Dividend =Divisor × Quotient + Remainder
Let us divide and equate the obtained remainder with (x + a)
Now, it is given that p(x) when divided by x2 – 2x + k leaves (x + a) as remainder.
Let us equate the remainder with x + a (as given in the question)
(-9 + 2k)x + 10 – 8k + k2 = x + a
Let us compare the coefficient of both LHS and RHS.
-9 + 2k = 1
⇒ 2k = 10
⇒ k = 5 —– (1)
Also, 10 – 8k + k2 = a —– (2)
As we obtained the value of k, let us substitute in equation (2) to find the value of a.
a = 10 – 40 + 25
a = – 5
Therefore, the value of k is 5 and a is – 5.
Q 6 – Find all the zeroes of the polynomial x4 – 3x3 + 6x – 4, if two of its zeroes are √2 and – √2
Given: Zeroes of the polynomial x4 – 3x3 + 6x – 4 is √2 and -√2
To find: All zeroes of the given polynomial.
let , p(x) = x4 – 3x3 + 6x – 4
From the given zeroes and using factor theorem,
( x – √2 ) and ( x + √2 ) are the factors of p(x)
⇒ x² – 2 is a factor of p(x)
Now, on dividing p(x) with x² – 2
we get,
p (x) = ( x² – 2 )( x² – 3x + 2 )
= ( x – √2 )( x + √2 ) ( x² – 2x – x + 2 )
= ( x – √2 )( x + √2 ) [ x ( x – 2 ) – ( x – 2 ) ]
= ( x – √2 )( x + √2 )( x – 2 )( x – 1 )
Q 7 – If α and β are the zeroes of the polynomial ax2 + bx + c, find the value of α2 + β2 .
ax2 + bx + c = 0
Q 8 – If α and β are the zeroes of a polynomial such that α + β = – 6 and αβ = 5, then find the polynomial.
We have been given α and β are the zeroes of the polynomial such that α + β= 6 and α β = 4 .
Sum of Zeroes:
α + β = 6
Product of Zeroes :
α β = 4
General Formula of Quadratic Polynomial:
f (x) = x² – ( α + β ) x + αβ
Find the Quadratic Polynomial:
x² – ( α + β ) x + αβ
x² – ( 6 ) x + 4
x² – 6x + 4
Therefore, Required Polynomial will be x² – 6x + 4.
Q 9 – If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x)?
By, Division Algorithm, we have
p (x) = g (x) q (x) + r (x)
If, on division of p(x) by g(x) the quotient is 0, then the degp (x)< degg (x)
Ex: p(x) = x +1 and g (x) = x2 −1
Here, the quotient is 0 and the degp(x) < degg (x) ie 1< 2
Q 10 – Find the zeroes of the quadratic polynomial 9t2 – 6t + 1 and verify the relationship between the zeroes and the coefficients.
9t2 – 6t + 1 = 0
9t2 – 3t -3t + 1 = 0
3t (3t -1) -1 (3t-1) =0
(3t-1) (3t-1) = 0
Both roots are same and = t = 1/3.
Sum of the roots = -b/a = -(-6/9) = 2/3
1/3 + 1/3 = 2/3
Product of roots = (c/a) = 1/9
1/3 x 1/3 = 1/9.
Q 11 – Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2.
Q 12 – If one of the zeros of the quadratic polynomial (k – 1)x2 + kx + 1 is -3, then find the value of k.
Q 13 – Find the zeros of the polynomial p(x) = 4x – 12x + 9.
4x² – 12x – 9 = 0
4x² – 6x – 6x-9 = 0
2x(2x–3) – 3 (2x–3) = 0
(2x-3) (2x-3) = 0
(2x-3)² = 0
2x – 3 = 0
2x = 3
x = 3/2
Q 14 – If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1, the remainder comes out to be (ax + b), find a and b.
Concept: 1 Mark
Application: 1 Mark
Calculation: 2 Marks
Let us divide the polynomial f(x)= 6x4 + 8x3 + 17x2 + 21x + 7 by the polynomial g(x)= 3x2 + 4x + 1
Clearly remainder x + 2
It is given that remainder is ax + b
ax + b = x + 2
Clearly ,remainder = x + 2
It is given that the remainder is ax + b
ax + b = x + 2
⇒ = 1 ,b = 2 [ on comparing the coefficients of like powers of x]
Q 15 – If (–1 and 2) are two zeroes of the polynomial 2x3 – x2 – 5x – 2, find its third zero.
x =–1/2
Let the third zeroes of the polynomial x
the two zeroes are given -1 and 2
so
(x + 1) (x – 2) = x2 – 2x + x – 2
x2 – x – 2
by using remainder theorem’
after dividing
Quatient = 2x + 1 =0
2x = – 1
x = –1/2
Q 16 – If 2 and -3 are the zeroes of the quadratic polynomial x2 + (a + 1) x + b; then find the values of a and b.
Ans. If -2 and 3 are the zeros of the quadratic polynomial
p(x)=x2 + (a+1) x + b
then,
p (−2) = (−2)2 + (a+1)(−2) + b = 0
= 4−2a−2+b = 0
= 2 −2a +b = 0
= 2a – b = 2
= a = (2+b) /2 ………
P (3) = (3)2 + (a+1) (3) + b = 0
= 9 + 3a + 3 + b = 0
3a + b +12 = 0….. (2)
put the value of a from (1) into (2)
3 (2+b) / 2 + b + 12 = 0
6 +3b + 2b + 24 = 0 5 b + 30 = 0b = −30/5 = −6
put the value of b in (1)
a = (2−6)/2 = −4/2 = − 2
Q 17 – If one zero of the polynomial 2x2 + 3x + λ is 1/2 find the value of and other zero.
p(x) = 2x² + 3x + k
One zero is ½
p(½) = 2(½)² + 3(½) + k = 0
2(¼) + 3/2 + k = 0
½+3/2 + k = 0
k + 4/2 = 0
k+2 = 0
k = –2
Therefore, given polynomial is 2x² + 3x – 2
Finding another zero:-
Let another zero be x
Sum of zeroes = –3/2
½ + x = – 3/2
x = –3/2 – ½
x = –4/2
x = – 2
(OR)
Product of zeroes = – 2/2
x(½) = – 1
x/2 = – 1
x = – 1×2
x = – 2
Therefore, another zero = – 2
Q 18 – Find the zeros of the polynomial f(x) = x3 – 5x2 – 2x + 24, if it is given that the product of its two zeros is 12.
f(x) = x³ – 5x² – 2x + 24
And we have given that the product of it’s two zeroes is 12 .
So, let the zeroes of the given cubic polynomial be φ , β and γ .
From the given condition we have,
φβ = 12 ………………(1)
and also we have ,
φ + β + γ = coefficient of x²/coefficient of x³ = 5 ………………….(2)
φβγ = – constant term/ coefficient of x³ = -24 ………………………(3)
Substituting the value of φβ in equation no. (3) we get,
φβγ = – 24
12γ = – 24
γ = – 24/12
γ =– 2 ………………….(4)
Putting the value of γ = -2 in equation no. (2) we get
φ + β + γ = 5
φ + β + (– 2) = 5
φ + β = 5 +2
φ + β = 7 ……………….(5)
Now, squaring on both sides we will get ,
(φ + β)² = (7)²
We know the expression of quadratic polynomial [ (φ + β)² = (φ – β)²+ 4φβ) ]
∴ ( φ – β )² + 4 × 12 = 49 [∵ φβ = 12 ]
(φ – β)² + 48 = 49
( φ – β)² = 49 – 48
(φ – β)² = 1
∴ φ – β = 1 ……………..(6)
Now, we will add the equation no. (5) and equation no. (6) we get,
φ + β = 7
φ – β = 1
—————-
2φ = 8
φ = 8/2
φ = 4
Putting φ = 4 in equation no. (5) we get,
φ – β = 1
4 – β = 1
– β = 1 – 4
-β = – 3 …….(multiplying by -1 on both side)
β = 3
∴ The zeroes are φ, β, γ = 4, 3, – 2
Q 19 – If one zero of polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, find the value of a.
If one of the zero is reciprocal of the other, let one zero be α
other zero will be 1/α
Product of roots = c/a
⇒ α × 1/α = 6a/(a²+9)
⇒ 1 = 6a/(a²+9)
⇒ a² + 9 = 6a
⇒ a² – 6a + 9 = 0
⇒ a² – 3a – 3a + 9= 0
⇒ a(a-3) -3(a-3) = 0
⇒ (a-3)(a-3) = 0
⇒ (a-3)²= 0
⇒ a = 3
Q 20 – Divide 2x4 – 9x3 + 5x2 + 3x – 8 by x2 – 4x+ 1 and verify the division algorithm.
In dividing the given factors,
Hence, Quotient
Remainder = -7
In division algorithm the quotient and divisor is multiplied and added with the remainder which results in the dividend. This method is usually used to verify if the quotient and remainder we obtained are right.
Verification of the division algorithms,
Quotient Divisor + Remainder
(2x2 – x -1) (x2 – 4x + 1) + (-7)
(2x2 – x -1)
Q 21 – On dividing the polynomial 4x4 – 5x3 – 39x2 – 46x – 2 by the polynomial g(x), the quotient and remainder were x2 – 3x – 5 and -5x + 8 respectively. Find g(x).
Given polynomial p(x) = 4x4 – 5x3 – 39x2 – 46x – 2
Given Quotient q(x) = x2 – 3x – 5
Given Remainder r(x) = -5x + 8.
We know that g(x) = p(x) – r(x)/q(x)
= 4x4 – 5x3 – 39x2 – 46x – 2 – (-5x + 8)/x2 – 3x – 5
= 4x4 – 5x3 – 39x2 – 46x – 2 + 5x – 8/x2 – 3x – 5
= 4x4 – 5x3 – 39x2 – 41x – 10/x2 – 3x – 5.
Now,
4x2 + 7x + 2
—————————————————–
x2 – 3x – 5) 4x4 – 5x3 – 39x2 – 41x – 10
4x4 – 12x3 – 20x2
———————————————————–
7x3 – 19x2 – 41x
7x3 – 21x2 – 35x
—————————————————————
2x2 – 6x – 10
2x2 – 6x – 10
——————————————————————– 0.
Hence, g(x) = 4x2 + 7x + 2.
Q 22 – What must be subtracted or added to p(x) = 8 x4 + 14x3 – 2x2 + 8x – 12 so that 4x2 + 3x – 2 is a factor of p(x)?
Polynomial to be subtracted is 15x − 14.
Q 23 – On dividing 3x3 + 4x2 + 5x – 13 by a polynomial g(x) the quotient and remainder were 3x +10 and 16x – 43 respectively. Find the polynomial g(x).
Given that,
let Polynomial + 4x2+ 5x−13
quotient
remainder
Now,
we know that
Euclid’s division lemma theorem,
+ 4x2+ 5x −13 = g (x) × (3x +10) + (16x−43)
+4x2+5x−13−16x+43 = g (x)×(3x+10)
+4x2−11x+30=g(x)×(3x+10)
now, dividing,
3x+10)3x3+4×2−11x+30(x2−2x+3
−3x3+10x2
_____________
−6x2−11x
−6x2
−20x
_____________
9x + 30
9x + 30
____________
0
Hence, g(x)=x2−2x + 3
Q 24 – What must be subtracted from p(x) = 8x4 + 14x3 – 2x2 + 7x – 8 so that the resulting polynomial is exactly divisible by g(x) = 4x2 + 3x – 2?
We know that
Dividend = Quotient × Divisor + Remainder
⇒ Dividend – Remainder = Quotient × Divisor
Clearly, RHS of the above result is divisible by the divisor.
Therefore, LHS is also divisible by the divisor. Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisor.
Dividing 8x4 + 14x3 − 2x2 + 7x−8 by 4x2+3x−2, we get
∴ Quotient =2x2 + 2x −1 and Remainder =14x − 10
Thus, if we subtract the remainder 14x −10 from 8x4 + 14x3 − 2x2 + 7x− 8, it will be divisible by 4x2 + 3x − 2.
Q 25 – If the polynomial f(x) = x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a. Find k and a.
By division algorithm, we have It is given that f(a) = x4 – 6x3 + 16x2 – 25x + 10, when divided by x – 2x + k leaves x + a as remainder.
:. f(x) – (x + a) = x4 – 6x3 + 16x2 – 25x + 10 – a is is exactly divisible by x – 2x + k.
Let us now divide x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k
For f(x) – (x + a) = x4 – 6x3 + 16x2 – 25x + 10 – a to be exactly divisible by x2 – 2x + k, we must have
(–10 + 2k) x + (10 – a – 8k + k2 ) = 0 for all x
⇒ – 10 + 2k = 0, 10 – a – 8k + k2 = 0
⇒ k = 5, 10 – a – 40 + 25 = 0
⇒ k = 5 and a = – 5
Q 26 – Find the zeros of the polynomial f(x) = x3 – 12x2 + 39x – 28, if it is given that the zeros are in AP.
Let , α = a-d , β = a and γ = a + d be the zeroes of the polynomial.
f ( x ) = x³ – 12x² + 39x – 28.
Therefore , α + β + γ = – [ – 12 / 1 ] = 12
And, αβγ = – [ – 28 / 1 ] = 28.
( a – d ) + a + ( a + d ) = 12
And ( a – d ) a ( a + d ) = 28.
3a = 12 and a ( a² – d² ) = 28
a = 4 and 4 ( 16 – d² ) = 28
a = 4 and 16 – d² = 7
a = 4 and d² = 9
a= 4 and d = ± 3
∴ α = a-d = 4 – 3 = 1
β = 4 (value of a )
and γ = a + d = 7
Final answers :
α = 7
β = 4
And,
γ = 1
Hence, the zeroes of the given polynomial f ( x ) = x³ – 12x² + 39x – 28 are 1 , 4 and 7.
Q 27 – If and β are the zeros of the quadratic polynomial f (x) = ax2 + bx + c, then evaluate: 
Q 28 – If the zeros of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are in A.P., prove that 2b3 – 3abc + a2d = 0.
Q 29 – If two zeros of the polynomial f(x) = x4 – 6x3 – 26x2 + 138x – 35 are 2 ,find other zeros.
The two zeroes of the polynomial is 3,2− 3
Therefore, √ 3)(x−2− √3)=x2+4−4x−3
= −4x+1 is a factor of the given polynomial.
√Using division algorithm, we get
− 6x3 −26x2+138x − 35 = (x2−4x+1)(x2−2x−35)
So, −2x−35) is also a factor of the given polynomial.
−2x−35=x2−7x+5x−35
Q 30 – Given that x – √5 is a factor of the cubic polynomial x3 – 3 √5x2 + 13x – 3√5 , find all the zeroes of the polynomial.
p(x) = x³ – 3√5x² + 13x – 3√5
x-√5 is a root of p(x)
We will divide p(x) by x-√5,
x-√5 ) x³ – 3√5x² + 13x – 3√5( x² – 2√5x + 3
x³ – √5x²
(-) (+)
-2√5x² + 13x
-2√5x² + 10x
(+) (-)
3x – 3√5
3x – 3√5
(-) (+)
0
Now we get a quadratic equation as quotient , we will find the roots of the quotient :
x² – 2√5x + 3
Roots = -b ± √(b²- 4ac)
2a
= -(-2√5) ± √[(-2√5)² – 4*1*3]
2*1
= 2√5 ± √(20 – 12)
2
= 2√5 ± √8
2
= 2√5 ± 2√2
2
√5 + √2 and √5 – √2
All the roots are √5 , √5 + √2 and √5 – √2.