Polynomials - NCERT Questions

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.

**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.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) *x*^{2} - 2 *x* - 8

(ii) 4*s*^{2} - 4*s* + 1

(iii) 6*x*^{2} - 3 - 7*x*

(iv) 4*u*^{2} + 8*u*

(v) *t*^{2} - 15

(vi) 3*x*^{2} - *x* - 4

**Soln. :** (i) We have *p*(*x*) = *x*^{2} - 2*x* - 8

= *x*^{2} + 2*x* - 4*x* - 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 *x*^{2} - 2*x* - 8 are 4 and - 2

Now, sum of the zeroes =

Thus, the relationship between zeroes and the coefficients in *x*^{2} - 2*x* - 8 is verified.

(ii)We have *p*(*s*) = 4*s*^{2} - 4*s* + 1

= 4*s*^{2} - 2*s* - 2*s* + 1

= 2*s*(2*s* - 1) - 1(2*s* - 1)

= (2*s* - 1)(2*s* - 1)

Sum of the zeroes

Sum of the zeroes

Sum of the zeroes

Sum of the zeroes

Thus, the relationship between the zeroes and coefficients in the polynomial 4*s*^{2} - 4*s* + 1 is verified.

(iii)We have *p*(*x*) = 6*x*^{2} - 3 - 7*x*

= 6*x*^{2} - 7*x* - 3

= 6*x*^{2} - 9*x* + 2*x* - 3

= 3*x*(2*x* - 3) + 1(2*x* - 3)

= (3*x* + 1)(2*x* - 3)

For *p*(*x*) = 0, we have,

Thus, the relationship between the zeroes and coefficients in the polynomial 6*x*^{2} - 3 - 7*x* is verified.

(iv)We have, *f*(*u*) = 4*u*^{2} + 8*u* = 4*u*(*u* + 2)

For *f*(*u*) = 0,

Either4*u* = 0 ⇒ *u* = 0

or *u* + 2 = 0 ⇒ *u* = -2

∴ The zeroes of 4*u*^{2} + 8*u* are 0 and - 2.

Now, 4*u*^{2} + 8*u* can be written as 4*u*^{2} + 8*u* + 0.

Thus, the relationship between zeroes and the coefficients in the polynomial 4*u*^{2} + 8*u* is verified.

(v)

For *f*(*t*) = 0, we have

Thus, the relationship between zeroes and the coefficients in the polynomial *t*^{2} - 15 is verified.

(vi) We have, *f*(*x*) = 3*x*^{2} - *x* - 4

= 3*x*^{2} + 3*x* - 4*x* - 4

= 3*x*(*x* + 1) - 4(*x* + 1)

= (*x* + 1)(3*x* - 4)

For *f*(*x*) = 0 ⇒ (*x* + 1)(3*x* - 4) = 0

Either(*x* + 1) = 0 ⇒ *x* = -1

Thus, the relationship between the zeroes and coefficients in 3x^{2} - x - 4 is verified

Find a quadratic polynomial each with the given numbers as sum and product of its zeroes respectively.

(i)

(ii)

(iii)

(iv) 1, 1

(v)

(vi) 4, 1

**Soln. :** (i) Since, sum of the zeroes,

Product of the zeroes, αβ = - 1

∴ The required quadratic polynomial is
*x*^{2} - (α + β)*x* + αβ

=

Since, and (4*x*^{2} - *x* - 4) have same zeroes, therefore (4*x*^{2} - *x* - 4) is the required quadratic polynomial.

(ii) Since, sum of the zeroes,

Product of zeroes,

∴ The required quadratic polynomial is

*x*^{2} - (α + β)*x* + αβ

Since, and have same zeroes, therefore is the required quadratic polynomial.

(iii) Since, sum of zeroes, (α + β) = 0

Product of zeroes,

∴ The required quadratic polynomial is

*x*^{2} - (α + β)*x* + αβ* *

(iv) Since, sum of zeroes, (α + β) = 1

Product of zeroes, αβ = 1

∴ The required quadratic polynomial is

*x*^{2} - (α + β)*x* + αβ

= *x*^{2} - (1)*x* + 1 = *x*^{2} - *x* + 1

(v) Since, sum of the zeroes, Product of zeroes,

∴ The required quadratic polynomial is

*x*^{2} - (α + β)*x* + αβ

Since, and have same zeroes, therefore, the required quadratic polynomial is

(vi) Since, sum of zeroes, (α + β) = 4 and product of zeroes, αβ = 1

∴ The required quadratic polynomial is

*x*^{2} - (α + β)*x* + αβ = *x*^{2} - 4*x* + 1

Divide the polynomial *p*(*x* ) by the polynomial *g*(*x* ) and find the quotient and remainder in each of the following :

(i) *p*(*x* ) = *x*^{3} - 3 *x*^{2} + 5 *x* - 3, *g*(*x* ) = *x*^{2} - 2

(ii) *p*(*x* ) = *x*^{4} - 3 *x*^{2} + 4 *x* + 5, *g*(*x* ) = *x*^{2} + 1 - *x*

(iii) *p*(*x* ) = *x*^{4} - 5 *x* + 6, *g*(*x* ) = 2 - *x*^{2}

**Soln. :** (i) Here, dividend *p*(*x*) = *x*^{3} - 3*x*^{2} + 5*x* - 3, divisor *g*(*x*) = *x*^{2} - 2

∴ We have

Thus, the quotient = (*x* - 3)

and remainder = (7*x* - 9)

(ii) Here, dividend *p*(*x*) = *x*^{4} - 3*x*^{2} + 4*x* + 5

and divisor *g*(*x*) = *x*^{2} + 1 - *x* = *x*^{2} - *x* + 1

∴ We have

Thus, the quotient is (*x*^{2} + *x* - 3) and remainder = 8

(iii) Here, dividend, *p*(*x*) = *x*^{4} - 5*x* + 6 and divisor, *g*(*x*) = 2 - *x*^{2} = - *x*^{2} + 2

∴ We have

Thus, the quotient = -*x*^{2} - 2

and remainder = -5*x* + 10.

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :

(i) *t*^{2} - 3; 2 *t*^{4} + 3 *t*^{3} - 2 *t*^{2} - 9 *t* - 12

(ii) *x*^{2} + 3 *x* + 1; 3 *x*^{4} + 5 *x*^{3} - 7 *x*^{2} + 2 *x* +2

(iii) *x*^{3} - 3 *x* + 1; *x*^{5} - 4 *x*^{3} + *x*^{2} + 3 *x* + 1

**Soln. :** (i) Dividing 2*t*^{4} + 3*t*^{3} - 2*t*^{2} - 9*t* - 12 by *t*^{2} - 3, we have

Remainder = 0

∴ (*t*^{2} - 3) is a factor of 2*t*^{4} + 3*t*^{3} - 2*t*^{2} - 9*t* - 12.

(ii) Dividing 3*x*^{4} + 5*x*^{3} - 7*x*^{2} + 2*x* + 2 by *x*^{2} + 3*x* + 1, we have

Remainder = 0

∴ *x*^{2} + 3*x* + 1 is a factor of 3*x*^{4} + 5*x*^{3} - 7*x*^{2} + 2*x* + 2.

(iii) Dividing *x*^{5} - 4*x*^{3} + *x*^{2} + 3*x* + 1 by *x*^{3} - 3*x* + 1, we get

The remainder = 2, *i.e*., remainder ≠ 0

∴ *x*^{3} - 3*x* + 1 is not a factor of *x*^{5} - 4*x*^{3} + *x*^{2}

+ 3*x* + 1.

Obtain all other zeroes of 3 *x*^{4} + 6 *x*^{3} - 2 *x*^{2} - 10 *x* - 5, if two of its zeroes are and -.

**Soln. :** We have *p*(*x*) = 3*x*^{4} + 6*x*^{3} - 2*x*^{2} - 10*x* - 5.

Given and - are zeroes of *p*(*x*).

∴ is a factor of p(x).

Now, let us divide 3*x*^{4} + 6*x*^{3} - 2*x*^{2} - 10*x* - 5 by

.

∴ 3*x*^{4} + 6*x*^{3} - 2*x*^{2} - 10*x* - 5

= (3*x*^{2} + 6*x* +3) (*x*^{2} - 5/3)

= 3(*x*^{2} + 2*x* + 1)

= 3(*x* + 1)^{2 }(*x*^{2} - 5/3)

Thus, the other zeroes of the given polynomial are -1 and -1.

On dividing *x*^{3} - 3 *x*^{2} + *x* + 2 by a polynomial *g*(*x* ), the quotient and remainder were *x* - 2 and -2 *x* + 4 respectively. Find *g*(*x* ).

**Soln. :** Here, Dividend, *p*(*x*) = *x*^{3} - 3*x*^{2} + *x* + 2

Divisor = *g*(*x*),

Quotient = (*x* - 2) and Remainder = (-2*x* + 4)

Since, (Quotient × Divisor) + Remainder

= Dividend

∴ [(*x* - 2) × *g*(*x*)] + [(-2*x* + 4)] = *x*^{3} - 3*x*^{2} + *x* + 2

⇒ (*x* - 2) × *g*(*x*) = *x*^{3} - 3*x*^{2} + *x* + 2 - (-2*x* + 4)

= *x*^{3} - 3*x*^{2} + *x* + 2 + 2*x* - 4

= *x*^{3} - 3*x*^{2} + 3*x* - 2

∴ *g*(*x*) =

Now, dividing *x*^{3} - 3*x*^{2} + 3*x* - 2 by *x* - 2, we have

∴ *g*(*x*) = = *x*^{2} - *x* + 1

Thus, the required divisor *g*(*x*) = *x*^{2} - *x* + 1.

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

**Soln. :** (i) *p*(*x*) = 3*x*^{2} - 6*x* + 27,

*g*(*x*) = 3 and *q*(*x*) = *x*^{2} - 2*x* + 9.

Now, deg *p*(*x*) = deg *q*(*x*)

*r*(*x*) = 0

⇒ *p*(*x*) = *q*(*x*) × *g*(*x*) + *r*(*x*).

(ii) *p*(*x*) = 2*x*^{3} - 2*x*^{2} + 2*x* + 3,

*g*(*x*) = 2*x*^{2} - 1, *q*(*x*) = *x* - 1 and

*r*(*x*) = 3*x* + *2*, deg *q*(*x*) = deg *r*(*x*)

⇒ *p*(*x*) = *q*(*x*) × *g*(*x*) + *r*(*x*)

(iii) *p*(*x*) = 2*x*^{3} - 4*x*^{2} + *x* + 4,

*g*(*x*) = 2*x*^{2} + 1, *q*(*x*) = *x* - 2 and

*r*(*x*) = 6, deg *r*(*x*) = 0

⇒ *p*(*x*) = *q*(*x*) × *g*(*x*) + *r*(*x*)

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 *x*^{3} + *x*^{2} - 5 *x* + 2; , 1, - 2

(ii) *x*^{3} - 4 *x*^{2} + 5 *x* - 2; 2, 1, 1

**Soln. :** ∴ (i) *p*(*x*) = 2*x*^{3} + *x*^{2} - 5*x* + 2

⇒ 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*) = 2*x*^{3} + *x*^{2} - 5*x* + 2

∴ Comparing it with *ax*^{3} + *bx*^{2} + *cx* + *d*, we have *a* = 2, *b* = 1, *c* = - 5 and *d* = 2

Also , 1 and - 2 are the zeroes of *p*(*x*).

and product of zeroes = αβγ

Thus, the relationship between the coefficients and the zeroes of *p*(*x*) is verified.

(ii) Here, *p*(*x*) = *x*^{3} - 4*x*^{2} + 5*x* - 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*) = *x*^{3} - 4*x*^{2} + 5*x* - 2 with *ax*^{3} + *bx*^{2} + *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*

Product of zeroes = αβγ = (2)(1)(1) = 2 = -*d*/*a*

Thus, the relationship between the zeroes and the coefficients of *p*(*x*) is verifed.

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 *ax*^{3} + *bx*^{2} + *cx* + *d* and its zeroes be α, β and γ.

If *a* = 1, then

and

∴ The required cubic polynomial is

1*x*^{3} + (-2)*x*^{2} + (-7)*x* + 14

= *x*^{3} - 2*x*^{2} - 7*x* + 14.

If the zeroes of the polynomial *x*^{3} - 3 *x*^{2} + *x* + 1 are *a* - *b* , *a* , *a* + *b* , find *a* and *b* .

**Soln. :** We have *p*(*x*) = *x*^{3} - 3*x*^{2} + *x* + 1.

Comparing it with *Ax*^{3} + *Bx*^{2} + *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*)

⇒ (*a* - *b*) + *a* + (*a* + *b*) = 3

⇒ 3*a* = 3

⇒ *a* = 1

⇒ (*a* - *b*) × *a* × (*a* + *b*) = -1

⇒ (1 - *b*) × 1 × (1 + *b*) = -1

⇒ 1 - *b*^{2} = -1

⇒ *b*^{2} = 1 + 1 = 2

Thus, *a* = 1 and

If two zeroes of the polynomial *x*^{4} - 6 *x*^{3} - 26 *x*^{2} + 138 *x* - 35 are find other zeroes.

**Soln. :** Here, *p*(*x*) = *x*^{4} - 6*x*^{3} - 26*x*^{2} + 138*x* - 35

∵ Two of the zeroes of *p*(*x*) are :

= (*x*^{2} + 4 - 4*x*) - 3 = *x*^{2} - 4*x* + 1

or *x*^{2} - 4*x* + 1 is a factor of *p*(*x*).

Now, dividing *p*(*x*) by *x*^{2} - 4*x* + 1, we have

∴ (*x*^{2} - 4*x* + 1)(*x*^{2} - 2*x* - 35) = *p*(*x*)

⇒ (*x*^{2} - 4*x* + 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.

If the polynomial *x*^{4} - 6 *x*^{3} + 16 *x*^{2} - 25 *x* + 10 is divided by another polynomial *x*^{2} - 2 *x* + *k* , the remainder comes out to be ( *x* + *a* ), find *k* and *a* .

**Soln. :** Applying the division algorithm to the polynomials *x*^{4} - 6*x*^{3} + 16*x*^{2} - 25*x* + 10 and *x*^{2} - 2*x* + *k*, we have

∴ Remainder = (2*k* - 9)*x* - *k*(8 - *k*) + 10

But the remainder = *x* + *a*

Therefore, comparing them, we have

2*k* - 9 = 1 ⇒ 2*k* = 1 + 9 = 10

⇒

and *a* = - *k*(8 - *k*) + 10

= - 5(3) + 10 = - 15 + 10 = - 5

Thus, *k* = 5 and *a* = - 5