Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3

Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3

Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

1. p(x) = x³ – 3x² + 5x – 3; g(x) = x² – 2
2. p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
3. p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Solution:
1. By applying the division algorithm to the given polynomials p(x) and g(x)

∴ Remainder = 7x – 9
degree of 7x – 9 = 1 < degree of (x² – 2) = 2.
⇒ Quotient = x – 3
Remainder = 7x – 9
Verify: Quotient × Division + Remainder
= (x – 3) (x² – 2) + (7x – 9)
= x³ – 2x – 3x² + 6 + 7x – 9
= x³ – 3x² + 5x – 3 = Dividend.

2. p(x) = x⁴ – 3x² + 4x + 5
g(x) = x² + 1 – x
Standard form of g(x) = x² – x + 1.
Now, by applying the division algorithm to the given polynomials p(x) and g(x)

degree of 8 = 0 < degree of x² – x + 1 = 2
So, Quotient = x² + x – 3
Remainder = 8.
Verify: Quotient × Divisor + Remainder
= (x² + x – 3) (x² – x + 1) + 8
= x⁴ – x³ + x² + x³ – x² + x – 3x² + 3x – 3 + 8
= x⁴ – 3x² + 4x + 5
= Dividend

3. p(x) = x⁴ – 5x + 6
g(x) = 2 – x²
Standard form of g(x) = -x² + 2
Now, by applying division algorithm to the given polynomial p(x) and g(x).

∵ deg. of -5x + 10 = 1 < deg. of -x² + 2 = 2
So, Quotient = -x² – 2
Remainder = -5x + 10.
Verify: Quotient × Divisor + Remainder
= (-x² – 2) × (-x² + 2) + (-5x + 10)
= (-x²)² – 4 – 5x + 10
= x⁴ – 5x + 6 = Dividend

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

1. t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
2. x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
3. x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:
By applying division algorithm to the given polynomials,

Since, the remainder is 0, hence the first polynomial is a factor of the second polynomial.

2. x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
By applying division algorithm to the given polynomials,

Since the remainder is 0, hence the first polynomial is a factor of the second polynomial.

3. x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
By applying division algorithm to the given polynomials,

Since the remainder is 2 (≠0), hence the first polynomial is not a factor of second polynomial.

So, 3x⁴ + 6x³ – 2x² – 10x – 5
= [x² – 5/3] (3x² + 6x + 3)
Now, 3x² + 6x + 3
= 3x² + 3x + 3x + 3
= 3x(x + 1) + 3(x + 1)
= 3(x + 1) (x + 1)
So, its zeroes are -1 and -1.
⇒ Remaining two zeroes of the given polynomial are -1 and -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
p(x) = x³ – 3x² + x + 2
q(x) = x – 2
r(x) = -2x + 4
By division algorithm for polynomials
p(x) = g(x) × q(x) + r(x)
⇒ x³ – 3x² + x + 2
= (x – 2) g(x) + (-2x + 4)
⇒ (x – 2) g(x)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
∴ g(x) = (x³ – 3x² + 3x – 2) ÷ (x – 2)

∴ g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

1. deg p(x) = deg q(x)
2. deg q(x) = deg r(x)
3. deg r(x) =0.

Solution:
1. deg p(x) = deg q(x)
Let p(x) = 8x² – 8x + 16
and g(x) = 8
∴ q(x) = x² – x + 2
r(x) = 0
p(x) = g(x) × q(x) + r(x)

2. deg q(x) = deg r(x)
Let p(x) = x³ + x² + x + 1
and g(x) = x² – 1
∴ q(x) = x + 1
r(x) = 2x + 2
p(x) = g(x) × q(x) + r(x)

3. deg r(x) = 0
Let p(x) = x⁴ + x³ + x² – 2x – 3
and g(x) = x² – 2
∴ q(x) = x² + x + 3
r(x) = 3
p(x) = g(x) × q(x) + r(x)

Follow on Facebook page – Click Here

Google News join in – Click Here

Read More Asia News – Click Here

Read More Sports News – Click Here

Read More Crypto News – Click Here