Short Answer Type Questions (Part - 1) - Polynomials Class 10 Notes | EduRev

Q 1. Find the sum and product of zeroes of 3x² - 5x + 6.

Here, p (x) = 3x² - 5x + 6
Comparing it with ax² + bx + c, we have
a = 3, b = - 5, c = 6
∴ Sum of the zeroes =

and, Product of the zeroes =

Q 2. Find the sum and product of the zeroes of polynomial p (x) = 2x³ - 5x_² - 14x + 8.

Comparing p (x) = 2x³ - 5x² - 14x + 8 with ax³ + bx² + cx + d, we have
a = 2, b = –5,
c = - 14 and d = 8
∴ Sum of the zeroes =
Product of zeroes

Q 3. Find a Quadratic polynomial whose zeroes are .

Sum of zeroes (S)

Product of roots (P)


Since the required Quadratic polynomial
= k(x² - Sx + P) ; where k is any real number.

= k

Thus, the required polynomial is
= k (x² - 2x - 1/4)

Q 4. If α and β are the zeroes of a Quadratic polynomial x² + x - 2 then find the value of .

Comparing x² + x - 2 with ax² + bx + c, we have:
a = 1, b = 1, c = - 2

Thus,

Q 5. If a and b are the zeroes of x² + px + q then find the value of .

Comparing x² + px + q with ax² + 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

Q 6. Find the zeroes of the quadratic polynomial 6x² - 3 - 7x.

We have,
= 6x² - 3 - 7x = 6x² - 7x - 3
= 6x² - 9x + 2x - 3
= 3x (2x - 3) + 1 (2x - 3)
= (3x + 1) (2x - 3)
For 6x² - 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.

Q 7. Find the zeroes of 2x² - 8x + 6.

We have,
2x² - 8x + 6 = 2x² - 6x - 2x + 6
= 2x (x - 3) - 2 (x - 3)
= (2x - 2) (x - 3)
= 2 (x - 1) (x - 3)
For 2x² - 8x + 6 to be zero,
Either, x - 1 = 0 ⇒ x = 1
or x - 3 = 0 ⇒ x = 3
∴ The zeroes of 2x² - 8x + 6 are 1 and 3.

Q 8. Find the zeroes of the quadratic polynomial 3x² + 5x - 2.

We have,
p (x) = 3x² + 5x - 2
= 3x² + 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 3x² + 5x - 2 are - 2 and 1/3.

Q 9. If the zero of a polynomial p (x) = 3x² - px + 2 and g (x) = 4x² - q x - 10 is 2, then find the value of p and q.

∵ p (x) = 3x² - 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) = 4x² - 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.

Q 10. Find the value of ‘k’ such that the quadratic polynomial 3x² + 2kx + x - k - 5 has the sum of zeroes as half of their product.

Here, p (x) = 3x² + 2kx + x - k - 5
= 3x² + (2k + 1) x - (k + 5)
Comparing p (x) with ax² + 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

Q 11. On dividing p (x) by a polynomial x - 1 - x², the Quotient and remainder were (x - 2) and 3 respectively. Find p (x).

Here,dividend = p (x)
Divisor, g (x) = (x - 1 - x²)
Quotient, q(x) = (x - 2)
Remainder, r (x) = 3
∵ Dividend = [Divisor × Quotient] +Remainder
∴ p (x)= [g (x) × q(x)] + r (x)
= [(x - 1 - x²) (x - 2)] + 3
= [x² - x - x³ - 2x + 2 + 2x²] + 3
= 3x² - 3x - x³ + 2 + 3
= - x³ + 3x² - 3x + 5

Q 12. Find the zeroes of the polynomial f (x) = 2 - x².

We have f (x)= 2 - x²
= (√2 )² - x²

Q 13. 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 (x² - 8x + 15) (x + 2)
= k (x³ - 8x² + 15x + 2x² - 16x + 30
= k (x³ + [- 8 + 2] x² + [15 - 16] x + 30)
= k (x³ - 6x² - x + 30)
Thus, the required polynomial is k (x³ - 6x² - x + 30).

Q 14. 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) = x³ - (a + b + γ) x² + (αβ + βγ + γα) x - (αβγ)
∴The required cubic polynomial is
= k {x³ - (3) x² + (- 10) x - (- 24)}
= k(x³ - 3x² - 10x + 24)

Note: If α, β and γ be the zeroes of a cubic polynomial p (x) then
p (x) = x³ - [Sum of the zeroes] x² + [Product of the zeroes taken two at a time] x - [Product of zeroes]
i.e., p (x) = k {x³ - (α + β + γ) x² + [αβ + βγ + γα] x - (αβγ).

Q 15. Find all the zeroes of the polynomial 4x⁴ - 20x³ + 23x² + 5x - 6 if two of its zeroes are 2 and 3.

Here, p (x) =4x⁴ - 20x³ + 23x² + 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)
⇒ x² - 5x + 6 is a factor of p (x)
Now, using the division algorithm for x² - 5x + 6 and the given polynomial p (x), we
get:

∴ We get (x² - 5x + 6) (4x² - 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: 

Q 16. If 1 is a zero of x³ - 3x² - x + 3 then find all other zeroes.

Here,p (x) = x³ - 3x² - 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) (x² - 2x - 3)
⇒ p (x) = (x - 1) [(x² - 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).

Q 17. Find all the zeroes of 2x⁴ - 3x³ - 3x² + 6x - 2, if two of its zeroes are 1 and 1/2.

Here, p(x) = 2x⁴ - 3x³ - 3x² + 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)
⇒ (2x² - 3x + 1) is a factor of p (x).
Now, dividing p (x) by 2x² - 3x + 1, we get

∴ p (x) = (2x² − 3x + 1) (x² − 2)

 are the zeroes of p (x).

Q 18. On dividing 4x³ - 8x² + 8x + 1 by a polynomial g (x), the Quotient and remainder were (2x² - 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.

Q 19. If α and β are the zeroes of the quadratic polynomial p (x) = kx² + 4x + 4 such that α² + β² = 24, find the value of k.

Here, p (x) = kx² + 4x + 4.
Comparing it with ax² + 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
∵ α² + β² = 24
∴ (α + β)² - 2αβ = 24
[∵ (x + y)² = x² + y² + 2xy ⇒ (x + y)² - 2xy = x² + y²]
⇒
⇒  
⇒ 16 − 8k − 24k² =0
⇒ 24k² + 8k − 16 = 0
⇒ (3k − 2) (k + 1) = 0
⇒ 3k − 2= 0 or k + 1 = 0
⇒ k = 2/3 or k = -1

Q 20. Find the zeroes of the quadratic polynomial 6x² - 3 - 7x and verify the relationship between the zeroes and the coefficients of the polynomial.

Here, p (x) = 6x² - 3 - 7x = 6x² - 7x - 3
= 6x² - 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.