Chapter 2: Polynomials

NCERT Solutions for Class 10th: Ch 2 Polynomials Maths8

Exercises 2.1

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.



Answer

(i) The number of zeroes is 0 as the graph does not cut thex-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects thex-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects thex-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects thex-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects thex-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects thex-axis at 3 points.
Page No: 33

Exercise 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i)x2– 2x– 8
(ii) 4s2– 4s+ 1
(iii) 6x2– 3 – 7x
(iv) 4u2+ 8u
(v)t2– 15
(vi) 3x2–x– 4

Answer

(i)x2– 2x– 8
= (x- 4) (x+ 2)
The value ofx2– 2x– 8 is zero whenx- 4 = 0 orx+ 2 = 0, i.e., whenx= 4 orx= -2
Therefore, the zeroes ofx2– 2x– 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient ofx)/Coefficient ofx2
Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient ofx2

(ii) 4s2– 4s+ 1
= (2s-1)2
The value of 4s2- 4s+ 1 is zero when 2s- 1 = 0, i.e., s = 1/2
Therefore, the zeroes of 4s2- 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient ofs)/Coefficient ofs2
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient ofs2.

(iii) 6x2– 3 – 7x
=6x2– 7x– 3
= (3x+ 1) (2x- 3)
The value of 6x2- 3 - 7xis zero when 3x+ 1 = 0 or 2x- 3 = 0, i.e.,x= -1/3 orx= 3/2
Therefore, the zeroes of 6x2- 3 - 7xare -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient ofx)/Coefficient ofx2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient ofx2.

(iv) 4u2+ 8u
=4u2+ 8u +0
= 4u(u+ 2)
The value of 4u2+ 8uis zero when 4u= 0 oru+ 2 = 0, i.e.,u= 0 oru= - 2
Therefore, the zeroes of 4u2+ 8uare 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient ofu)/Coefficient ofu2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient ofu2.

(v)t2– 15
=t2- 0.t- 15
= (t- √15) (t+ √15)
The value oft2- 15 is zero whent- √15= 0 ort+ √15= 0, i.e., whent= √15ort= -√15
Therefore, the zeroes oft2- 15 are √15and -√15.Sum of zeroes = √15+ -√15= 0 = -0/1 = -(Coefficient oft)/Coefficient oft2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient ofu2.

(vi) 3x2–x– 4
= (3x- 4) (x+ 1)
The value of 3x2–x– 4 is zero when 3x- 4 = 0 andx+ 1 = 0,i.e., whenx= 4/3 orx= -1
Therefore, the zeroes of 3x2–x– 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient ofx)/Coefficient ofx2
Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient ofx2.

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , -1
(ii) √2, 1/3
(iii) 0, √5
(iv) 1,1
(v) -1/4 ,1/4
(vi) 4,1

Answer

(i) 1/4 , -1
Let the polynomial beax2+bx+c, and its zeroes be α and ß
α + ß = 1/4 = -b/a
αß = -1 = -4/4 =c/a
Ifa= 4, thenb= -1,c= -4
Therefore, the quadratic polynomial is 4x2-x-4.

(ii) √2, 1/3
Let the polynomial beax2+bx+c, and its zeroes be α and ß
α + ß = √2= 3√2/3 = -b/a
αß = 1/3 =c/a
Ifa= 3, thenb= -3√2,c= 1
Therefore, the quadratic polynomial is 3x2-3√2x+1.

(iii) 0, √5
Let the polynomial beax2+bx+c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a
αß = √5= √5/1 =c/a
Ifa= 1, thenb= 0,c= √5
Therefore, the quadratic polynomial isx2+ √5.

(iv) 1, 1
Let the polynomial beax2+bx+c, and its zeroes be α and ß
α + ß = 1= 1/1 = -b/a
αß = 1= 1/1 =c/a
Ifa= 1, thenb= -1,c= 1
Therefore, the quadratic polynomial isx2-x+1.

(v) -1/4 ,1/4
Let the polynomial beax2+bx+c, and its zeroes be α and ß
α + ß = -1/4= -b/a
αß = 1/4 =c/a
Ifa= 4, thenb= 1,c= 1
Therefore, the quadratic polynomial is 4x2+x+1.

(vi) 4,1
Let the polynomial beax2+bx+c, and its zeroes be α and ß
α + ß = 4 = 4/1= -b/a
αß = 1= 1/1 =c/a
Ifa= 1, thenb= -4,c= 1
Therefore, the quadratic polynomial isx2- 4x+1.

Page No: 36

Exercise 2.3

1. Divide the polynomialp(x) by the polynomialg(x) and find the quotient and remainder in each of the following:
Answer

(i)p(x) =x3– 3x2+ 5x– 3,g(x) =x2– 2

Quotient =x-3 and remainder 7x- 9

(ii)p(x) =x4– 3x2+ 4x + 5,g(x) =x2+ 1 –x

Quotient =x2+x- 3 and remainder 8

(iii)p(x) =x4– 5x+ 6,g(x) = 2 –x2

Quotient = -x2-2 and remainder -5x+10

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

Answer
(i)t2– 3,  2t4+ 3t3– 2t2– 9t– 12

t2– 3 exactly divides  2t4+ 3t3– 2t2– 9t– 12 leaving no remainder. Hence, it is a factor of  2t4+ 3t3– 2t2– 9t– 12.

(ii)x2+ 3x+ 1, 3x4+ 5x3– 7x2+ 2x+ 2

x2+ 3x+ 1 exactly divides 3x4+ 5x3– 7x2+ 2x+ 2 leaving no remainder. Hence, it is factor of 3x4+ 5x3– 7x2+ 2x+ 2.

(iii)x3– 3x+ 1,x5– 4x3+x2+ 3x+ 1

x3– 3x+ 1 didn't divides exactlyx5– 4x3+x2+ 3x+ 1 and leaves 2 as remainder. Hence, it not a factor ofx5– 4x3+x2+ 3x+ 1.

3. Obtain all other zeroes of 3x4+ 6x3– 2x2– 10x– 5, if two of its zeroes are √(5/3)
and - √(5/3).

Answer

p(x) = 3x4+ 6x3– 2x2– 10x– 5
Since the two zeroes are √(5/3)and - √(5/3).

We factorizex2+ 2x+ 1
= (x+ 1)2
Therefore, its zero is given byx+ 1 = 0
x= -1
As it has the term (x+ 1)2, therefore, there will be 2 zeroes atx= - 1.
Hence, the zeroes of the given polynomial are √(5/3)and - √(5/3), - 1 and - 1.

4.  On dividingx3- 3x2+x+ 2 by a polynomialg(x), the quotient and remainder werex- 2 and
-2x+ 4, respectively. Findg(x).

Answer

Here in the given question,
Dividend =x3- 3x2+x+ 2
Quotient =x- 2
Remainder = -2x+ 4
Divisor =g(x)
We know that,
Dividend = Quotient × Divisor + Remainder
⇒x3- 3x2+x+ 2 = (x- 2) ×g(x) + (-2x+ 4)⇒x3- 3x2+x+ 2 - (-2x+ 4) = (x- 2) ×g(x)
⇒x3- 3x2+ 3x- 2 = (x- 2) ×g(x)
⇒g(x) =  (x3- 3x2+ 3x- 2)/(x- 2)


∴g(x) = (x2-x+ 1)

5.Give examples of polynomialp(x),g(x),q(x) andr(x), which satisfy the division algorithm and
(i) degp(x) = degq(x)
(ii) degq(x) = degr(x)
(iii) degr(x) = 0

Answer

(i) Let us assume the division of 6x2+ 2x+ 2 by 2
Here,p(x) = 6x2+ 2x+ 2
g(x) = 2
q(x) = 3x2+x+ 1
r(x) = 0
Degree ofp(x) andq(x) is same i.e. 2.
Checking for division algorithm,
p(x) =g(x) ×q(x) +r(x)
Or, 6x2+ 2x+ 2 = 2x(3x2+x+ 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division ofx3+xbyx2,
Here,p(x) =x3+x
g(x) =x2
q(x) =xandr(x) =x
Clearly, the degree ofq(x) andr(x) is the same i.e., 1.
Checking for division algorithm,
p(x) =g(x) ×q(x) +r(x)
x3+x= (x2) ×x+x
x3+x=x3+x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division ofx3+ 1 byx2.
Here,p(x) =x3+ 1
g(x) = x2
q(x) =xandr(x) = 1
Clearly, the degree ofr(x) is 0.
Checking for division algorithm,
p(x) =g(x) ×q(x) +r(x)
x3+ 1 = (x2) ×x+ 1
x3+ 1 =x3+ 1
Thus, the division algorithm is satisfied.

Page No: 36

Exercise 2.4 (Optional)
1. 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) 2x3+x2- 5x+ 2; 1/2, 1, -2
(ii)x3- 4x2+ 5x -2; 2, 1, 1

Answer

(i)p(x) = 2x3+x2- 5x+ 2
Now for zeroes, putting the given value in x.

p(1/2) = 2(1/2)3+(1/2)2- 5(1/2)+ 2
= (2×1/8) + 1/4 - 5/2 + 2
= 1/4 + 1/4 - 5/2 + 2
= 1/2 - 5/2 + 2 = 0

p(1) = 2(1)3+(1)2- 5(1)+ 2
= (2×1) + 1 - 5 + 2
= 2 + 1 - 5 + 2 = 0

p(-2) = 2(-2)3+ (-2)2- 5(-2)+ 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3+ bx2+ cx+ d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2+ 1 - 2
⇒ 1/2 = 1/2
c/a =αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2

-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)p(x) =x3- 4x2+ 5x -2
Now for zeroes, putting the given value in x.

p(2) = 23- 4(2)2+ 5(2)- 2
= 8 - 16 + 10 - 2
= 0

p(1) = 13- 4(1)2+ 5(1)- 2
= 1 - 4 + 5 - 2
= 0

p(1) = 13- 4(1)2+ 5(1)- 2
= 1 - 4 + 5 - 2
= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax3+ bx2+ cx+ d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2+ 1 + 1
⇒ 4 = 4
c/a =αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5

-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Answer

Let the polynomial beax3+bx2+cx + dand the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will bex3- 2x2-7x + 14

3. If the zeroes of the polynomialx3– 3x2+x+ 1 are a–b, a, a+b, find a and b.

Answer

Since, (a - b), a, (a + b) are the zeroes of the polynomialx3– 3x2+x+ 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a2- ab + a2+ ab + a2- b2= 1
⇒ 3a2- b2=1

Putting the value of a,

⇒ 3(1)2- b2= 1
⇒ 3 - b2= 1
⇒ b2= 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2
4. If two zeroes of the polynomial x4– 6x3– 26x2+ 138x – 35 are  2±√3,  find other zeroes.

Answer

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x4– 6x3– 26x2+ 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x2- 4x + 4 = 3,
⇒ x2- 4x + 1= 0

Now, dividing p(x) by x2- 4x + 1


∴ p(x) = x4- 6x3- 26x2+ 138x - 35
= (x2- 4x + 1) (x2- 2x - 35)
= (x2- 4x + 1) (x2- 7x + 5x - 35)
= (x2- 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x2- 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.

5. If the polynomial 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.

Answer

On dividing x4– 6x3+ 16x2– 25x + 10 by x2– 2x + k



∴ Remainder = (2k - 9)x - (8 - k)k + 10
But the remainder is given as x+ a.

On comparing their coefficients,
2k - 9 = 1
⇒ k = 10
⇒ k = 5 and,

-(8-k)k + 10 = a
⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5

Hence, k = 5 and a = -5

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