# NCERT Solutions for Class 10 Maths

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1.     (i)   The graph does not meets x-axis at all. Hence, it does not have any zero.

(ii)  Graph meets x-axis 1 time. It means this polynomial has 1 zero.

(iii) Graph meets x-axis 3 times. Therefore, it has 3 zeroes.

(iv)  Graph meets x-axis 2 times. Therefore, it has 2 zeroes.

(v)   Graph meets x-axis 4 times. It means it has 4 zeroes.

(vi)  Graph meets x-axis 3 times. It means it has 3 zeroes.

2.    (i)  x– 2x – 8

Comparing given polynomial with general form ax2+bx+c,

We get a = 1, b = -2 and c = -8

We have, x– 2x – 8

= x– 4x + 2x – 8

= x(x – 4) + 2 (x – 4) = (x – 4)(x + 2)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

(x-4)(x+2) = 0

⇒ x = 4, -2 are two zeroes.

(ii)  4s– 4s + 1

Here, a = 4, b = -4 and c = 1

We have, 4s– 4s + 1

= 4s– 2s – 2s + 1 = 2s(2s – 1) -1(2s – 1)

= (2s – 1)(2s – 1)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

⇒ (2s – 1)(2s – 1) = 0

⇒

Therefore, two zeroes of this polynomial are

(iii) 6x– 3 – 7x

Here, a = 6, b = -7 and c = -3

We have, 6x– 3 – 7x

= 6x– 7x – 3 = 6x– 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3) = (2x – 3)(3x + 1)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

⇒       (2x-3)(3x+1) = 0

Therefore, two zeroes of this polynomial are,

(iv) 4u2+8u

Here,  a = 4, b = 8 and c = 0

4u2+8u = 4u(u+2)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

⇒ 4u(u+2) = 0

⇒ u = 0,-2

Therefore, two zeroes of this polynomial are 0, -2

(v)  t– 15

Here, a= 1, b = 0 and c = -15

We have, t– 15 ⇒ t2= 15⇒ t = ±√15

Therefore, two zeroes of this polynomial are √15, -√15

(vi) 3x– x – 4

Here, a = 3, b = -1 and c = -4

We have,  3x– x – 4 = 3x– 4x + 3x – 4

= x(3x – 4) +1(3x – 4) = (3x – 4) (x+1)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

⇒ (3x-4)(x+1) = 0

⇒

Therefore, two zeroes of this polynomial are

3.    (i)

Let a and β are two zeroes of above quadratic polynomial.

∴ Quadratic polynomial which satisfies above conditions = 4x– x – 4

(ii)

Let a and β be two zeros of above quadratic polynomial.

∴ Quadratic polynomial which satisfies above conditions = 3x– 3√2x +1

(iii) 0, √5

Let a and β be two zeros of above quadratic polynomial.

∴ a = 1, b = 0, c = √5

∴ Quadratic polynomial which satisfies above conditions = x2+ √5

(iv) 1, 1

Let a and β be two zeros of above quadratic polynomial.

∴ a = 1, b = -1, c = 1

∴ Quadratic polynomial which satisfies above conditions = x– x + 1

(v)

Let quadratic polynomial be ax+ bx + c

Let a and β be two zeros of above quadratic polynomial.

∴ a = 4, b =1, c =1

∴ Quadratic polynomial which satisfies above conditions = 4x+ x + 1

(vi) 4, 1

Let a and β be two zeros of above quadratic polynomial.

∴ Quadratic polynomial which satisfies above conditions = x– 4x + 1

4.   (i)

Therefore, quotient = x – 3 and Remainder = 7x – 9

(ii)

Therefore, quotient = x2 + x – 3 and, Remainder = 8

(iii)

Therefore, quotient = -x2 – 2 and, Remainder = -5x + 10

5.     (i)

∵ Remainder = 0

Hence first polynomial is a factor of second polynomial.

(ii)

∵ Remainder = 0

Hence first polynomial is a factor of second polynomial.

(iii)

∵ Remainder ≠ 0

Hence first polynomial is not factor of second polynomial.

6.

Applying Division Algorithm to find more factors we get:

We have p(x) = g(x) x q(x).

Therefore, other two zeroes of (3x+ 6x– 2x– 10x -5) are -1 and -1.

7.     Let p(x) = x– 3x2+x+2, q(x) = (x – 2) and r(x) = (-2x+4)

According to Polynomial Division Algorithm, we have

p(x) = g(x).q(x) + r(x)                        ⇒  x3-3x2+x+2 = g(x).(x-2)-2x+4

⇒ x3-3x2+x+2+2x-4 = g(x). (x-2)   x3-3x2+3x-2 = g(x). (x-2)

8.    (i)

So, we can see in this example that deg p(x) = deg q(x) = 2

(ii)

We can see in this example that deg q(x) = deg r(x) = 1

(iii) Let p(x) = x2+5x-3, g(x) = x+3

We can see in this example that deg r(x) = 0

9.    Comparing the given polynomial with ax3 + bx2 + cx + d , we get a=2,b=1,c=-5 and d = 2.

(ii) Comparing the given polynomial with ax3 + bx2 + cx + d , we get

a =1, b = -4, c = 5 and d = -2.

p(2) = 2(23-4(22+5(2)-2 = 8-16+10-2 =0

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

∴ 2, 1 and 1 are the zeroes of x34x2 +5x – 2.

10.   Let the cubic polynomial be ax3 + bx2 + cx + d and its zeroes be a, β and y.

Here, a =1, b = -2, c = -7 and d = 14

Hence, cubic polynomial will be x32x27x +14.

11.   Since (a – b) , a ,(a + b) are the zeroes of the polynomial x3 -3x2 + 3x +1.

12.    Since 2 ± √3 are two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 +138x – 35.

Let x = 2 ± √3             ⇒    x – 2 =±√3

Squaring both sides, ⇒    x2 – 4x+ 4 =3               ⇒          x2 – 4x +1 = 0

Now we divide p (x) by x2 – 4x +1 to obtain other zeroes.

∴  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)1 = (x2 -4x+1)(x+5)(x-7)

⇒ (x + 5) and (x-7) are the other factors of p(x).

∴ -5 and 7 are other zeroes of the given polynomial.

13.   Let us divide x4 — 6.x3 +16x2 — 25x +10 by x2 — 2x + k.

∴ Remainder = (2k —9)x—(8—k)k +10

On comparing this remainder with given remainder, i.e. x+ a,

2k – 9 = 1   ⇒   2k =10     ⇒      k = 5

And -(8 – k)k + 10 =a     ⇒       a = -(8 – 5) 5 + 10= -5

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