Chapter 2 – Polynomials (Ex – 2.3)

Polynomials

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) = r³ – 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. Here, dividend and divisor are both in standard forms. So, we have:

2. Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as
x² – x + 1.
We have:

∴ The quotient is x² + x – 3 and the remainder is 8.

3. To carry out the division, we first write divisor in the standard form.
So, divisor = – x² + 2
We have:

∴ The quotient is – x² – 2 and the remainder is – 5x + 10.

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:
1. Let us divide 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3.
We have:

Since the remainder is 0, therefore t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

2. Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1.
We get,

Since, the remainder is 0, therefore x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2

3. Let us divide x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1. We get,

Here, remainder is 2(≠ 0). Therefore, x³ – 3x + 1 is not a factor of
x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all the zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are
√5/3 and – √5/3.
Solution:
Since two zeroes are √5/3 and – √5/3, so (x – √5/3) and (x + √5/3) are the factors of the given polynomial.
Now, (x – √5/3) (x + √5/3) = x² – 5/3.
So, (3x² – 5) is a factor of the given polynomial.
Applying the division algorithm to the given polynomial and 3x² – 5, we have:

∴ 3x⁴ + 6x³ – 2x² – 10x – 5
= (3x² – 5)(x² + 2x + 1)
Now, x² + 2x + 1
= x² + x + x + 1
= x(x + 1) + 1(x + 1)
= (x + 1)(x + 1)
So, its other zeroes are – 1 and – 1.
Thus, all the zeroes of the given fourth degree polynomial are √5/3, – √5/3, – 1 and – 1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quoitent and remainder were x – 2 and – 2x + 4 respectively. Find g(x).
Solution:
Since on dividing x³ – 3x² + x + 2 by a polynomial g(x), the quoitent and remainder were (x – 2) and (- 2x + 4) respectively, therefore
Quoitent × Divisor + Remainder = Dividend
or (x – 2) × g(x) + (- 2x + 4)
= x³ – 3x² + x + 2
or (x – 2) × g(x)
= x³ – 3x² + x + 2 + 2x – 4
or g(x)
= (x³−3x²+3x−2)/x−2 …………… (1)
Let us divide x³ – 3x² + 3x – 2 by x – 2. We get

∴ (1) gives 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:
There can be several examples for each of (i), (ii) and (iii).
However, one example for each case may be taken as under:

1. p(x) = 14, g(x)= 2, q(x) = x² – x + 7, r(x) = 0
2. p(x) = x³ + x² + x + 1, g(x) = x² – 1, q(x) = x + 1, r(x) = 2x + 2
3. p(x) = x³ + 2x² – x + 2, g(x) = – 1, q(x) = x + 2, r(x) = 4