Q1: Find the sum and product of zeroes of 3x^{2}  5x + 6.
Here, p (x) = 3x^{2}  5x + 6
Comparing it with ax^{2} + bx + c, we have
a = 3, b =  5, c = 6
∴ Sum of the zeroes =and, Product of the zeroes =
Q2: Find the sum and product of the zeroes of polynomial p (x) = 2x^{3}  5x^{2}  14x + 8.
Comparing p (x) = 2x^{3}  5x^{2}  14x + 8 with ax^{3} + bx^{2} + cx + d, we have
a = 2, b = –5,
c =  14 and d = 8
∴ Sum of the zeroes =
Product of zeroes
Q3: Find a Quadratic polynomial whose zeroes are .
Sum of zeroes (S)
Product of roots (P)
Since the required Quadratic polynomial
= k(x^{2}  Sx + P) ; where k is any real number.= k
Thus, the required polynomial is
= k (x^{2}  2x  1/4)
Q4: If α and β are the zeroes of a Quadratic polynomial x^{2} + x  2 then find the value of .
Comparing x^{2} + x  2 with ax^{2} + bx + c, we have:
a = 1, b = 1, c =  2
Thus,
Q5: If a and b are the zeroes of x^{2} + px + q then find the value of .
Comparing x^{2} + px + q with ax^{2} + bx + c
a =1, b = p and c = q
∴ Sum of zeroes, a + b =  b/a
⇒
and αβ = c/a
⇒ αβ = q/1 = q
Now,Thus, the value of is
Q6: Find the zeroes of the quadratic polynomial 6x^{2}  3  7x.
We have,
= 6x^{2}  3  7x = 6x^{2}  7x  3
= 6x^{2}  9x + 2x  3
= 3x (2x  3) + 1 (2x  3)
= (3x + 1) (2x  3)
For 6x^{2}  3  7x to be equal to zero,
either (3x + 1) = 0 or (2x  3) = 0
⇒ 3x =  1 or 2x = 3
⇒
Thus, the zeroes of and 3/2.
Q7: Find the zeroes of 2x^{2}  8x + 6.
We have,
2x^{2}  8x + 6 = 2x^{2}  6x  2x + 6
= 2x (x  3)  2 (x  3)
= (2x  2) (x  3)
= 2 (x  1) (x  3)
For 2x^{2}  8x + 6 to be zero,
Either, x  1 = 0 ⇒ x = 1
or x  3 = 0 ⇒ x = 3
∴ The zeroes of 2x^{2}  8x + 6 are 1 and 3.
Q8: Find the zeroes of the quadratic polynomial 3x^{2} + 5x  2.
We have,
p (x) = 3x^{2} + 5x  2
= 3x^{2} + 6x  x  2
= 3x (x + 2)  1 (x + 2)
= (x + 2) (3x  1)
For p (x) = 0, we get
Either x + 2 = 0 ⇒ x =  2
or 3x  1 = 0 ⇒ x = 1/3
Thus, the zeroes of 3x^{2} + 5x  2 are  2 and 1/3.
Q9: If the zero of a polynomial p (x) = 3x^{2}  px + 2 and g (x) = 4x^{2}  q x  10 is 2, then find the value of p and q.
∵ p (x) = 3x^{2}  px + 2
∴ p (2) = 3 (2)^{2}  p (2) + 2 = 0
[2 is a zero of p (x)]
or 12  2p + 2 or 14  2p = 0
or p = 7
Next g (x) = 4x^{2}  q x  10
∴ g (2) = 4(2)^{2}  Q (2)  10 = 0
[2 is a zero of g (x)]
or 4 × 4  2q  10 = 0
or 16  2q  10 = 0
or 6  2q = 0
⇒ q = 6/2 ⇒ q = 3
Thus, the required values are p = 7 and q = 3.
Q10: Find the value of ‘k’ such that the quadratic polynomial 3x^{2} + 2kx + x  k  5 has the sum of zeroes as half of their product.
Here, p (x) = 3x^{2} + 2kx + x  k  5
= 3x^{2} + (2k + 1) x  (k + 5)
Comparing p (x) with ax^{2} + bx + c, we have:
a = 3, b = (2k + 1),
c =  (k + 5)
∴ Sum of the zeroes
Product of the zeroes
According to the condition,
Sum of zeroes = 1/2 (product of roots)
⇒  2 (2k + 1) =  (k + 5)
⇒ 2 (2k + 1) = k + 5
⇒ 4k + 2 = k + 5
⇒ 4k  k = 5  2
⇒ 3k = 3
⇒ k = 3/3 = 1
Q11: On dividing p (x) by a polynomial x  1  x^{2}, the Quotient and remainder were (x  2) and 3 respectively. Find p (x).
Here,dividend = p (x)
Divisor, g (x) = (x  1  x^{2})
Quotient, q(x) = (x  2)
Remainder, r (x) = 3
∵ Dividend = [Divisor × Quotient] +Remainder
∴ p (x)= [g (x) × q(x)] + r (x)
= [(x  1  x^{2}) (x  2)] + 3
= [x^{2}  x  x^{3}  2x + 2 + 2x^{2}] + 3
= 3x^{2}  3x  x^{3} + 2 + 3
=  x^{3} + 3x^{2 } 3x + 5
Q12: Find the zeroes of the polynomial f (x) = 2  x^{2}.
We have f (x)= 2  x^{2}
= (√2 )^{2 } x^{2}
Q13: Find the cubic polynomial whose zeroes are 5, 3 and  2.
∵ 5, 3 and  2 are zeroes of p (x)
∴ (x  5), (x  3) and (x + 2) are the factors of p (x)
⇒ p (x) = k (x  5) (x  3) (x + 2)
= k (x^{2}  8x + 15) (x + 2)
= k (x^{3}  8x^{2} + 15x + 2x^{2}  16x + 30
= k (x^{3} + [ 8 + 2] x^{2} + [15  16] x + 30)
= k (x^{3 } 6x^{2}  x + 30)
Thus, the required polynomial is k (x^{3}  6x^{2}  x + 30).
Q14: If α, β and γ be the zeroes of a polynomial p (x) such that (α + β +γ) = 3, (αβ + βγ + γα) = 10 and αβγ =  24 then find p (x).
Here, α + β + γ = 3
αb + βγ + γα =  10
αβγ =  24
∵ A cubic polynomial having zeroes as α,β,γ is
p (x) = x^{3}  (a + b + γ) x^{2 }+ (αβ + βγ + γα) x  (αβγ)
∴The required cubic polynomial is
= k {x^{3}  (3) x^{2} + ( 10) x  ( 24)}
= k(x^{3 } 3x^{2}  10x + 24)Note: If α, β and γ be the zeroes of a cubic polynomial p (x) then
p (x) = x^{3}  [Sum of the zeroes] x^{2} + [Product of the zeroes taken two at a time] x  [Product of zeroes]
i.e., p (x) = k {x^{3}  (α + β + γ) x^{2} + [αβ + βγ + γα] x  (αβγ).
Q15: Find all the zeroes of the polynomial 4x^{4}  20x^{3} + 23x^{2} + 5x  6 if two of its zeroes are 2 and 3.
Here, p (x) =4x^{4}  20x^{3} + 23x^{2} + 5x  6
Since, 2 and 3 are the zeroes of p (x),
∴ (x  2) and (x  3) are the factors of p(x)
⇒ (x  2) (x  3) is a factor of p (x)
⇒ x^{2 } 5x + 6 is a factor of p (x)
Now, using the division algorithm for x^{2}  5x + 6 and the given polynomial p (x), we
get:∴ We get (x^{2}  5x + 6) (4x^{2}  1) = p (x)
⇒ (x  3) (x  2) [(2x)^{2}  (1)2] = p (x)
⇒ (x  3) (x  2) (2x  1) (2x + 1) = p (x)
⇒
Thus, all the zeroes of p (x) are:
Q16: If 1 is a zero of x^{3}  3x^{2}  x + 3 then find all other zeroes.
Here,p (x) = x^{3}  3x^{2}  x + 3
∵ 1 is a zero of p (x)
∴ (x  1) is a factor of p (x).
Now, dividing p (x) by (x  1), we have:∴ p (x) = (x  1) (x^{2}  2x  3)
⇒ p (x) = (x  1) [(x^{2}  3x + x  3)]
= (x  1) [x (x  3) + 1 (x  3)]
= (x  1) [(x  3) (x + 1)]
i.e., (x  1), (x  3) and (x + 1) are the factors of p (x).
⇒ 1, 3, and  1 are the zeroes of p (x).
Q17: Find all the zeroes of 2x^{4}  3x^{3}  3x^{2} + 6x  2, if two of its zeroes are 1 and 1/2.
Here, p(x) = 2x^{4}  3x^{3}  3x^{2} + 6x  2
∵ 1 and are the zeroes of p (x)
∴ (x  1) and are the factors of p (x)
⇒ (x  1) (2x  1) is a factor of p (x)
⇒ (2x^{2}  3x + 1) is a factor of p (x).
Now, dividing p (x) by 2x^{2}  3x + 1, we get∴ p (x) = (2x^{2} − 3x + 1) (x^{2} − 2)
are the zeroes of p (x).
Q18: On dividing 4x^{3}  8x^{2} + 8x + 1 by a polynomial g (x), the Quotient and remainder were (2x^{2}  3x + 2) and (x + 3) respectively. Find g(x).
∵ Dividend = Divisor × Quotient + Remainder
i.e., p (x) = g (x) × Q (x) + r (x)
∴ g (x) =Thus, the required polynomial g (x) = 2x  1.
Q19: If α and β are the zeroes of the quadratic polynomial p (x) = kx^{2} + 4x + 4 such that α^{2} + β^{2} = 24, find the value of k.
Here, p (x) = kx^{2} + 4x + 4.
Comparing it with ax^{2} + bx + c, we have:
a = k; b = 4; c = 4
∴ Sum of the zeroes = b/a
⇒ α + β = 4/k
and Product of the zeroes = c/a
⇒ αβ = 4/k
∵ α^{2} + β^{2 }= 24
∴ (α + β)^{2}  2αβ = 24
[∵ (x + y)^{2} = x^{2} + y^{2} + 2xy ⇒ (x + y)^{2}  2xy = x^{2} + y^{2}]
⇒
⇒
⇒ 16 − 8k − 24k^{2} =0
⇒ 24k^{2} + 8k − 16 = 0
⇒ (3k − 2) (k + 1) = 0
⇒ 3k − 2= 0 or k + 1 = 0
⇒ k = 2/3 or k = 1
Q20: Find the zeroes of the quadratic polynomial 6x^{2}  3  7x and verify the relationship between the zeroes and the coefficients of the polynomial.
Here, p (x) = 6x^{2}  3  7x = 6x^{2}  7x  3
= 6x^{2 } 9x + 2x  3
= 3x (2x  3) + 1 (2x  3)
= (2x  3) (3x + 1)
=
∴ Zeroes of p (x) are 3/2 and
To verify the relationship:
Sum of the zeroes =⇒
⇒
⇒ 7/6 = 7/6
L.H.S = R.H.S ⇒ Relationship is verified.
Product of the zeroes =⇒
⇒
i.e., L.H.S = R.H.S ⇒ Relationship is verified.
Q21: Find the zeroes of the quadratic polynomial 5x^{2}  4  8x and verify the relationship between the zeroes and the coefficients of the polynomial.
p (x) = 5x^{2}  4  8x
= 5x^{2}  8x  4
= 5x^{2}  10x + 2x  4
= 5x (x  2) + 2 (x  2)
= (x  2) (5x + 2)
∴ zeroes of p (x) are 2 and
Relationship Verification
Sum of the zeroes
⇒
⇒
⇒ 8/5 = 8/5i.e., L.H.S. = R.H.S. ⇒ relationship is verified.
Product of the zeroes =
⇒
⇒
i.e., L.H.S. = R.H.S.
⇒ The relationship is verified.
Q22: Find the quadratic polynomial, the sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.
The quadratic polynomial p (x) is given by
x^{2}  (Sum of the zeroes) x + (Product of the zeroes)
∴ The required polynomial is
= x^{2}  [8] x + [12]
= x^{2}  8x + 12
To find zeroes:
∵ x^{2}  8x + 12 = x^{2}  6x  2x + 12
= x (x  6)  2 (x  6)
= (x  6) (x  2)
∴ The zeroes of p (x) are 6 and 2.
Q23: If one zero of the polynomial (a^{2}  9) x^{2} + 13x + 6a is reciprocal of the other, find the value of ‘a’.
Here, p (x) = (a^{2}  9) x^{2} + 13x + 6a
Comparing it with Ax^{2} + Bx + C, we have:
A = (a^{2}  9); B = 13; C = 6
Let one of the zeroes = a
∴ The other zero = 1/α
Now, Product of the zeroes⇒
⇒ 6a = a^{2} − 9 ⇒ a^{2} − 6a + 9 = 0
⇒ (a − 3)^{2} =0 ⇒ a − 3=0
⇒ a = 3
Thus, the required value of a is 3.
Q24: If the product of zeroes of the polynomial ax^{2}  6x  6 is 4, find the value of ‘a’
Here, p (x) = ax^{2}  6x  6
∵ Product of zeroes =
but product of zeroes is given as 4
∴ ⇒ − 6 = 4 × a
⇒ ⇒Thus, the required value of a is 3/2.
Q25: Find all the zeroes of the polynomial x^{4} + x^{3}  34x^{2}  4x + 120, if two of its zeroes are 2 and  2.
Here p (x) = x^{4} + x^{3}  34x^{2}  4x + 120
∵ The two zeroes of p (x) are 2 and  2
∴ (x  2) and (x + 2) are factors of p (x)
⇒ (x  2) (x + 2) is a factor of p (x)
⇒ x^{2}  4 is a factor of p (x).
Now, dividing p(x) by x^{2}  4, we have:∵ Remainder = 0
∴ p (x) = (x^{2}  4) (x^{2} + x  30)
i.e., x^{2} + x  30 is also a factor of p (x).
∵ x^{2} + x  30 = x^{2} + 6x  5x  30 = x (x + 6)  5 (x + 6)
= (x + 6) (x  5) = [x  ( 6)] [x  5]
 6 and 5 are also zeroes of p (x).
⇒ All the zeroes of the given polynomial are : 2,  2, 5 and  6
Q26: Find all the zeroes of the polynomial 2x^{4} + 7x^{3}  19x^{2 } 14x + 30, if two of its zeroes are √2 and √2.
P(x) = 2x^{4} + 7x^{3}  19x^{2}  14x + 30
∵ √2 and √2 are the two zeroes of p (x).
∴ are the factors of p (x).
⇒ i.e., x^{2}  2 is a factor of p (x).
Now, dividing p (x) by x^{2}  2, we have:∴ p (x) = (2x^{2} + 7x  15) (x^{2}  2)
[∵ Remainder = 0]
⇒ 2x^{2} + 7x  15 is a factor of p (x)
∵ 2x^{2} + 7x  15 = 2x^{2} + 10x  3x  15
= 2x (x + 5)  3 (x + 5)
= (2x  3) (x + 5)
=
∴ 3/2 and  5 are zeroes of p (x)
are the zeroes of p (x).
Q27: Find the quadratic polynomial whose zeroes are 1 and  3. Verify the relation between the coefficients and the zeroes of the polynomial.
The given zeroes are 1 and  3.
∴ Sum of the zeroes = 1 + ( 3) =  2
Product of the zeroes = 1 × ( 3) =  3
A quadratic polynomial p (x) is given by
x^{2}  (sum of the zeroes) x + (product of the zeroes)
∴ The required polynomial is
x^{2 } ( 2) x + ( 3)
⇒ x^{2} + 2x  3
Verification of relationship
∵ Sum of the zeroes∴
⇒− 2= − 2i.e., L.H.S = R.H.S ⇒ The sum of zeroes is verified
∵ Product of the zeroes =
∴
⇒− 3= − 3i.e., L.H.S = R.H.S ⇒ The product of zeroes is verified.
Q28: Find the zeroes of the quadratic polynomial 4x^{2}  4x  3 and verify the relation between the zeroes and its coefficients.
Here, p (x) = 4x^{2}  4x  3 = 4x^{2}  6x + 2x  3
= 2x (2x  3) + 1 (2x  3)
= (2x  3) (2x + 1)
=
∴ are zeroes of p (x).Verification of relationship
∵ Sum of the zeroes =∴
⇒ 2/2 = 1 ⇒ 1= 1
⇒
2/2 = 1 ⇒ 1 = 1
i.e., L.H.S = R.H.S ⇒ Sum of zeroes is verified
Now, Product of zeroes =⇒
i.e., L.H.S = R.H.S ⇒ Product of zeroes is verified.
Q29: Obtain all other zeroes of the polynomial 2x^{3}  4x  x^{2} + 2, if two of its zeroes are √2 and √2.
p (x) = 2x^{3 } 4x  x^{2} + 2
∵ √2 and √2 are the zeroes of p (x)
∴ and are the factors of p (x)
⇒ is a factor of p (x)
⇒ x^{2}  2 is a factor of p (x)
Now, Dividing p (x) by (x^{2}  2), we have:⇒ p (x) = (x^{2}  2) (2x  1)
∴ 2x  1 is also a factor of p (x)
i.e.,is another factor of px.
⇒ 1/2 is another zero of p (x)
Q30: Find all the zeroes of x^{4}  3x^{3} + 6x  4, if two of its zeroes are √2 and  √2.
p (x) = x^{4 } 3x^{3} + 6x  4
∵ √2 and ( √2) are the zeroes of p(x)
∴ x√2 and x( √2) are factors of p (x)
⇒ is a factor of p (x).
⇒ x^{2}  2 is a factor of p (x)
On Dividing p (x) by x^{2 } 2, we have:Since, remainder = 0
∴ (x^{2}  2) (x^{2}  3x + 2) = p (x)
Now, x^{2}  3x + 2 = x^{2}  2x  x + 2
= x (x  2)  1 (x  2) = (x  1) (x  2)
i.e., (x  1) (x  2) is a factor of p (x)
∴ 1 and 2 are zeroes of p (x).
∴ All the zeroes of p (x) are ,√2 ,  √2, 1 and 2.
Q31: Find a quadratic polynomial whose zeroes are  4 and 3 and verify the relationship between the zeroes and the coefficients.
We know that:
P (x) = x^{2}  [Sum of the zeroes] x + [Product of the zeroes] ...(1)
∵ The given zeroes are  4 and 3
∴ Sum of the zeroes = ( 4) + 3 =  1
Product of the zeroes = ( 4) × 3 =  12
From (1), we have
x^{2}  ( 1) x + ( 12)
= x^{2} + x  12 ...(2)
Comparing (2) with ax^{2} + bx + c, we have
a = 1, b = 1, c =  12
∴ Sum of the zeroes = b/a
⇒ (+ 3) + ( 4) = 1/1
i.e., L.H.S = R.H.S ⇒ Sum of zeroes is verified.
Product of zeroes = c/a
⇒ 3 × ( 4) = 12/1
⇒  12 =  12
i.e., L.H.S = R.H.S ⇒ Product of roots is verified.
Q32: Using division algorithm, find the quotient and remainder on dividing f (x) by g (x), where f (x) = 6x^{3} + 13x^{2} + x  2 and g (x) = 2x + 1
Here, f (x) = 6x^{3} + 13x^{2} + x  2
g (x) = 2x + 1
Now, dividing f (x) by g (x), we have:Thus, The quotient = 3x2 + 5x  2
remainder = 0
Q33: If the polynomial 6x^{4} + 8x^{3} + 17x^{2} + 21x + 7 is divided by another polynomial 3x^{2} + 4x + 1 then the remainder comes out to be ax + b, find ‘a’ and ‘b’.
We have:
∴ Remainder = x + 2
Comparing x + 2 with ax + b, we have
a = 1 and b = 2
Thus, the required value of a = 1 and b = 2.
Q34: If the polynomial x^{4} + 2x^{3} + 8x^{2} + 12x + 18 is divided by another polynomial x^{2} + 5, the remainder comes out to be px + q. Find the values of p and q.
We have:
∴ Remainder = 2x + 3
Comparing 2x + 3 with px + q, we have
p = 2 and q = 3
Q35: Find all the zeroes of the polynomial x^{3} + 3x^{2}  2x  6, if two of its zeroes are  √2 and √2.
p (x) = x^{3} + 3x^{2}  2x  6
∵ Two of its zeroes are √2 and √2
⇒ is a factor of p (x)
⇒ x^{2}  2 is a factor of p (x).
Now, dividing p (x) by x^{2}  2 we have:∴ p (x) = (x^{2}  2) (x + 3)
i.e., (x + 3) is a factor of p (x),
⇒ ( 3) is a zero of p (x)
∴All the zeroes of p (x) are  √2, √2 and  3.
Q36: Find all the zeroes of the polynomial 2x^{3} + x^{2}  6x  3, if two of its zeroes are √3 and √3.
p (x) = 2x^{3} + x^{2}  6x  3
Two of its zeroes are √3 and √3
∴ and are factors of p (x)
i.e., is a factor of p (x)
⇒ x^{2}  3 is a factor of p (x)
Now, Dividing p (x) by x^{2}  3, we have:∴ p (x) = (x^{2}  3) (2x + 1)
⇒ is a factor of p (x)
⇒ 1/2 is a zero of p (x)
∴ All the zeroes of p (x) are √3 , √3 and 1/2.
Q37: Find the zeroes of the polynomial and verify the relation between the coefficients and the zeroes of the above polynomial.
The given polynomial is
∴ zeroes of the given polynomial are
Now in,
coefficient of x^{2} = 1
coefficient of x = 1/6
constant term = –2
∴ Sum of zeroesProduct of zeroes
Q38: Find the quadratic polynomial, the sum and product of whose zeroes are respectively. Also find its zeroes.
Sum of zeroes = √2
Product of zeroes
∵ A quadratic polynomial is given by
x^{2} – [sum of roots] x + [Product of roots]
∴ The required polynomial is
⇒
Since =
⇒ zeroes are
Q39: If the remainder on division of x^{3} + 2x^{2} + kx + 3 by x  3 is 21, then find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x^{3} + 2x^{2} + kx  18.
Let x^{3} + 2x^{2} + kx + 3 = p(x)
∵ The divisor = x – 3
∴ p(3) = 3^{3} + 2 × 3^{2} + 3k + 3
21 = 27 + 18 + 3k + 3
[∵ Remainder = 21]
⇒ 21 – 18 – 3 – 27 = 3k
⇒ –27 = 3k ⇒ k = – 9
Now, the given cubic polynomial
= x^{3} + 2x^{2} – 9x + 3
since,∴ The required quotient = x^{2} + 5x + 6
Now, x^{3} + 2x^{2} – 9x – 18 = (x – 3) (x^{2} + 5x + 6)
= (x – 3) (x + 3) (x + 2)
⇒ The zeroes of x^{3} + 2x^{2} – 9x – 18 are 3, –3 and – 2
Q40: If a and b are zeroes of the quadratic polynomial x^{2} – 6x + a; find the value of ‘a’ if 3α + 2β = 20.
We have quadratic polynomial = x^{2} – 6x + a ...(1)
∵ a and b are zeroes of (1)
∴
It is given that: 3α + 2β = 20 ...(2)
Now, α +β = 6 ⇒ 2 (α+ β) = 2(6)
2α + 2β = 12 ...(3)
Subtracting (3) from (2), we have
Substituting a = 8 in α + β= 6, we get
8 +β = 6 ⇒ β = –2
Since, αβ = a
8(–2) = α ⇒ α = –16
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