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Polynomials - NCERT Questions
Q 1.

The graphs of y = p(x) are given in the below figures, for some polynomials p(x ). Find the number of zeroes of p(x), in each case.
ncert

SOLUTION:

Soln. : (i) The given graph is parallel to x-axis, it does not intersect the x-axis.
∴ It has no zero.
(ii) The given graph intersects the x-axis at one point only.
∴ It has one zero.
(iii) The given graph intersects the x-axis at three points.
∴ It has three zeroes.
(iv) The given graph intersects the x-axis at two points.
∴ It has two zeroes.
(v) The given graph intersects the x-axis at four points.
∴ It has four zeroes.
(vi) The given graph meets the x-axis at three points.
∴ It has three zeroes.

Q 2.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 - 2 x - 8
(ii) 4s2 - 4s + 1
(iii) 6x2 - 3 - 7x
(iv) 4u2 + 8u
(v) t2 - 15
(vi) 3x2 - x - 4

SOLUTION:

Soln. : (i) We have p(x) = x2 - 2x - 8
= x2 + 2x - 4x - 8
= x(x + 2) - 4(x + 2)
= (x - 4)(x + 2)
For p(x) = 0, we have (x - 4)(x + 2) = 0
Either x - 4 = 0 ⇒ x = 4 or x + 2 = 0 ⇒ x = - 2
∴ The zeroes of x2 - 2x - 8 are 4 and - 2
Now, sum of the zeroes = ncert
ncert
Thus, the relationship between zeroes and the coefficients in x2 - 2x - 8 is verified.
(ii)We have p(s) = 4s2 - 4s + 1
= 4s2 - 2s - 2s + 1
= 2s(2s - 1) - 1(2s - 1)
= (2s - 1)(2s - 1)
ncert
Sum of the zeroes ncert
Sum of the zeroes ncert
Sum of the zeroes ncert
Sum of the zeroes ncert
Thus, the relationship between the zeroes and coefficients in the polynomial 4s2 - 4s + 1 is verified.
(iii)We have p(x) = 6x2 - 3 - 7x
= 6x2 - 7x - 3
= 6x2 - 9x + 2x - 3
= 3x(2x - 3) + 1(2x - 3)
= (3x + 1)(2x - 3)
For p(x) = 0, we have,
ncert
Thus, the relationship between the zeroes and coefficients in the polynomial 6x2 - 3 - 7x is verified.
(iv)We have, f(u) = 4u2 + 8u = 4u(u + 2)
For f(u) = 0,
Either4u = 0 ⇒ u = 0
or u + 2 = 0 ⇒ u = -2
∴ The zeroes of 4u2 + 8u are 0 and - 2.
Now, 4u2 + 8u can be written as 4u2 + 8u + 0.
ncert
Thus, the relationship between zeroes and the coefficients in the polynomial 4u2 + 8u is verified.
(v) ncert
For f(t) = 0, we have
ncert
Thus, the relationship between zeroes and the coefficients in the polynomial t2 - 15 is verified.
(vi) We have, f(x) = 3x2 - x - 4
= 3x2 + 3x - 4x - 4
= 3x(x + 1) - 4(x + 1)
= (x + 1)(3x - 4)
For f(x) = 0 ⇒ (x + 1)(3x - 4) = 0
Either(x + 1) = 0 ⇒ x = -1
ncert
Thus, the relationship between the zeroes and coefficients in 3x2 - x - 4 is verified

Q 3.

Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively.
(i) ncert
(ii) ncert
(iii) ncert
(iv) 1, 1
(v) ncert
(vi) 4, 1

SOLUTION:

Soln. : (i) Since, sum of the zeroes,
Product of the zeroes, αβ = - 1
∴ The required quadratic polynomial is
x2 - (α + β)x + αβ
= ncert
Since, and (4x2 - x - 4) have same zeroes, therefore (4x2 - x - 4) is the required quadratic polynomial.
(ii) Since, sum of the zeroes, ncert
Product of zeroes,
∴ The required quadratic polynomial is
x2 - (α + β)x + αβ
ncert
ncert
Since, and ncert have same zeroes, therefore ncert is the required quadratic polynomial.
(iii) Since, sum of zeroes, (α + β) = 0
Product of zeroes, ncert
∴ The required quadratic polynomial is
x2 - (α + β)x + αβ
ncert
(iv) Since, sum of zeroes, (α + β) = 1
Product of zeroes, αβ = 1
∴ The required quadratic polynomial is
x2 - (α + β)x + αβ
= x2 - (1)x + 1 = x2 - x + 1
(v) Since, sum of the zeroes, Product of zeroes,
∴ The required quadratic polynomial is
x2 - (α + β)x + αβ
ncert
ncert
Since, and ncert have same zeroes, therefore, the required quadratic polynomial is ncert
(vi) Since, sum of zeroes, (α + β) = 4 and product of zeroes, αβ = 1
∴ The required quadratic polynomial is
x2 - (α + β)x + αβ = x2 - 4x + 1

Q 4.

Divide the polynomial p(x ) by the polynomial g(x ) and find the quotient and remainder in each of the following :
(i) p(x ) = x3 - 3 x2 + 5 x - 3, g(x ) = x2 - 2
(ii) p(x ) = x4 - 3 x2 + 4 x + 5, g(x ) = x2 + 1 - x
(iii) p(x ) = x4 - 5 x + 6, g(x ) = 2 - x2

SOLUTION:

Soln. : (i) Here, dividend p(x) = x3 - 3x2 + 5x - 3, divisor g(x) = x2 - 2
∴ We have
ncert
Thus, the quotient = (x - 3)
and remainder = (7x - 9)
(ii) Here, dividend p(x) = x4 - 3x2 + 4x + 5
and divisor g(x) = x2 + 1 - x = x2 - x + 1
∴ We have
ncert
Thus, the quotient is (x2 + x - 3) and remainder = 8
(iii) Here, dividend, p(x) = x4 - 5x + 6 and divisor, g(x) = 2 - x2 = - x2 + 2
∴ We have
ncert
Thus, the quotient = -x2 - 2
and remainder = -5x + 10.

Q 5.

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
(i) t2 - 3; 2 t4 + 3 t3 - 2 t2 - 9 t - 12
(ii) x2 + 3 x + 1; 3 x4 + 5 x3 - 7 x2 + 2 x +2
(iii) x3 - 3 x + 1; x5 - 4 x3 + x2 + 3 x + 1

SOLUTION:

Soln. : (i) Dividing 2t4 + 3t3 - 2t2 - 9t - 12 by t2 - 3, we have
ncert
Remainder = 0
∴ (t2 - 3) is a factor of 2t4 + 3t3 - 2t2 - 9t - 12.
(ii) Dividing 3x4 + 5x3 - 7x2 + 2x + 2 by
x2 + 3x + 1, we have
ncert
Remainder = 0
x2 + 3x + 1 is a factor of 3x4 + 5x3 - 7x2 + 2x + 2.
(iii) Dividing x5 - 4x3 + x2 + 3x + 1 by
x3 - 3x + 1, we get
ncert
The remainder = 2, i.e., remainder ≠ 0
x3 - 3x + 1 is not a factor of x5 - 4x3 + x2
+ 3x + 1.

Q 6.

Obtain all other zeroes of 3 x4 + 6 x3 - 2 x2 - 10 x - 5, if two of its zeroes are ncertand -ncert.

SOLUTION:

Soln. : We have p(x) = 3x4 + 6x3 - 2x2 - 10x - 5.
Given ncert and -ncert are zeroes of p(x).
ncertis a factor of p(x).
Now, let us divide 3x4 + 6x3 - 2x2 - 10x - 5 by
ncert.

ncert
∴ 3x4 + 6x3 - 2x2 - 10x - 5
= (3x2 + 6x +3) (x2 - 5/3)
= 3(x2 + 2x + 1) ncert
= 3(x + 1)2 (x2 - 5/3)
Thus, the other zeroes of the given polynomial are -1 and -1.

Q 7.

On dividing x3 - 3 x2 + x + 2 by a polynomial g(x ), the quotient and remainder were x - 2 and -2 x + 4 respectively. Find g(x ).

SOLUTION:

Soln. : Here, Dividend, p(x) = x3 - 3x2 + x + 2
Divisor = g(x),
Quotient = (x - 2) and Remainder = (-2x + 4)
Since, (Quotient × Divisor) + Remainder
= Dividend
∴ [(x - 2) × g(x)] + [(-2x + 4)] = x3 - 3x2 + x + 2
⇒ (x - 2) × g(x) = x3 - 3x2 + x + 2 - (-2x + 4)
= x3 - 3x2 + x + 2 + 2x - 4
= x3 - 3x2 + 3x - 2
g(x) = ncert
Now, dividing x3 - 3x2 + 3x - 2 by x - 2, we have
ncert
g(x) = ncert = x2 - x + 1
Thus, the required divisor g(x) = x2 - x + 1.

Q 8.

Give examples of polynomials p(x ), g(x ), q(x ) and r(x ), which satisfy the division algorithm and
(i) deg p(x ) = deg q(x )
(ii) deg q(x ) = deg r(x )
(iii) deg r(x ) = 0

SOLUTION:

Soln. : (i) p(x) = 3x2 - 6x + 27,
g(x) = 3 and q(x) = x2 - 2x + 9.
Now, deg p(x) = deg q(x)
r(x) = 0
p(x) = q(x) × g(x) + r(x).
(ii) p(x) = 2x3 - 2x2 + 2x + 3,
g(x) = 2x2 - 1, q(x) = x - 1 and
r(x) = 3x + 2, deg q(x) = deg r(x)
p(x) = q(x) × g(x) + r(x)
(iii) p(x) = 2x3 - 4x2 + x + 4,
g(x) = 2x2 + 1, q(x) = x - 2 and
r(x) = 6, deg r(x) = 0
p(x) = q(x) × g(x) + r(x)

Q 9.

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
(i) 2 x3 + x2 - 5 x + 2; ncert, 1, - 2
(ii) x3 - 4 x2 + 5 x - 2; 2, 1, 1

SOLUTION:

Soln. : ∴ (i) p(x) = 2x3 + x2 - 5x + 2
ncert
ncert
ncert is a zero of p(x).
Again, p(1) = 2(1)3 + (1)2 - 5(1) + 2
= 2 + 1 - 5 + 2 = 0
⇒ 1 is a zero of p(x).
Also, p(-2) = 2(-2)3 + (-2)2 - 5(-2) + 2
= -16 + 4 + 10 + 2
= -16 + 16 = 0
⇒ - 2 is a zero of p(x).
Now, p(x) = 2x3 + x2 - 5x + 2
∴ Comparing it with ax3 + bx2 + cx + d, we have a = 2, b = 1, c = - 5 and d = 2
Also ncert, 1 and - 2 are the zeroes of p(x).
ncert
ncert ncert
ncert
ncert
and product of zeroes = αβγ
ncert
Thus, the relationship between the coefficients and the zeroes of p(x) is verified.
(ii) Here, p(x) = x3 - 4x2 + 5x - 2
p(2) = (2)3 - 4(2)2 + 5(2) - 2
= 8 - 16 + 10 - 2 = 18 - 18 = 0
⇒ 2 is a zero of p(x)
Again p(1) = (1)3 - 4(1)2 + 5(1) - 2
= 1 - 4 + 5 - 2 = 6 - 6 = 0
⇒ 1 is a zero of p(x).
Now, Comparing p(x) = x3 - 4x2 + 5x - 2 with ax3 + bx2 + cx + d = 0, we have
a = 1, b = - 4, c = 5 and d = - 2
2, 1 and 1 are the zeroes of p(x).
∴ Let α = 2, β = 1, γ = 1
Now, sum of zeroes = α + β + γ
= 2 + 1 + 1 = 4 = -b/a
ncert
Product of zeroes = αβγ = (2)(1)(1) = 2 = -d/a
Thus, the relationship between the zeroes and the coefficients of p(x) is verifed.

Q 10.

Find a cubic polynomial with the sum, sum of products of its zeroes taken two at a time and the product of its zeroes as 2, -7, -14 respectively.

SOLUTION:

Soln. : Let the required cubic polynomial be ax3 + bx2 + cx + d and its zeroes be α, β and γ.
ncert
ncert
ncert
If a = 1, then ncert
ncert
and ncert
∴ The required cubic polynomial is
1x3 + (-2)x2 + (-7)x + 14
= x3 - 2x2 - 7x + 14.

Q 11.

If the zeroes of the polynomial x3 - 3 x2 + x + 1 are a - b , a , a + b , find a and b .

SOLUTION:

Soln. : We have p(x) = x3 - 3x2 + x + 1.
Comparing it with Ax3 + Bx2 + Cx + D,
We have A = 1, B = -3, C = 1 and D = 1
It is given that (a - b), a and (a + b) are the zeroes of the polynomial.
∴ Let α = (a - b), β = a and γ = (a + b)
ncert
⇒ (a - b) + a + (a + b) = 3
⇒ 3a = 3
a = 1
ncert
⇒ (a - b) × a × (a + b) = -1
⇒ (1 - b) × 1 × (1 + b) = -1
⇒ 1 - b2 = -1
b2 = 1 + 1 = 2
Thus, a = 1 and ncert

Q 12.

If two zeroes of the polynomial
x4 - 6 x3 - 26 x2 + 138 x - 35 are ncert find other zeroes.

SOLUTION:

Soln. : Here, p(x) = x4 - 6x3 - 26x2 + 138x - 35
∵ Two of the zeroes of p(x) are : ncert
ncert
ncert
ncert
= (x2 + 4 - 4x) - 3 = x2 - 4x + 1
or x2 - 4x + 1 is a factor of p(x).
Now, dividing p(x) by x2 - 4x + 1, we have
ncert
∴ (x2 - 4x + 1)(x2 - 2x - 35) = p(x)
⇒ (x2 - 4x + 1)(x - 7)(x + 5) = p(x)
i.e., (x - 7) and (x + 5) are other factors of p(x).
∴ 7 and - 5 are other zeroes of the given polynomial.

Q 13.

If the polynomial x4 - 6 x3 + 16 x2 - 25 x + 10 is divided by another polynomial x2 - 2 x + k , the remainder comes out to be ( x + a ), find k and a .

SOLUTION:

Soln. : Applying the division algorithm to the polynomials x4 - 6x3 + 16x2 - 25x + 10 and x2 - 2x + k, we have
ncert
∴ Remainder = (2k - 9)x - k(8 - k) + 10
But the remainder = x + a
Therefore, comparing them, we have
2k - 9 = 1 ⇒ 2k = 1 + 9 = 10
ncert
and a = - k(8 - k) + 10
= - 5(3) + 10 = - 15 + 10 = - 5
Thus, k = 5 and a = - 5