Saturday, May 6, 2023

146-NCERT-10-2-Polynomials - Ex-2.3

NCERT
10th Mathematics
Exercise 2.3
Topic: 2 Polynomials

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EXERCISE 2.3

Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
(i) p(x) = x3 – 3x2 + 5x – 3,      g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5,     g(x) = x2 + 1 – x
(iii) p(x) = x4 – 5x + 6,              g(x) = 2 – x2

Explanation:

1) We know that Dividend = Divisor × Quotient + Remainder
2) If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find
polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x).

Solution:

(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Here quotient q(x) is (x - 3) and remainder r(x) is (7x - 9)
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Here quotient q(x) is (x2 + x - 3) and remainder r(x) is 8
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Here quotient q(x) is (- x2 - 2) and remainder r(x) is (- 5x + 10)

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t2 – 3,               2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1,     3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1,      x5 – 4x3 + x2 + 3x + 1

Explanation:

1) We know that Dividend = Divisor × Quotient + Remainder
2) If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find
polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x).
3) When the remainder is 0, then the polynomial g(x) is a factor of p(x).

Solution:

(i) t2 – 3,    2t4 + 3t3 – 2t2 – 9t – 12
1) Here first polynomial g(x) = t2 – 3.
2) Second polynomial p(x) = 2t4 + 3t3 – 2t2 – 9t – 12.
3) Divide the second polynomial by the first polynomial.
4) Here quotient is 2t2 + 3t + 4, and remainder is 0.
5) As the remainder is 0, (t2 – 3) is a factor of (2t4 + 3t3 – 2t2 – 9t – 12).

(ii) x2 + 3x + 1,     3x4 + 5x3 – 7x2 + 2x + 2
1) Here first polynomial g(x) = x2 + 3x + 1.
2) Second polynomial p(x) = 3x4 + 5x3 – 7x2 + 2x + 2.
3) Divide the second polynomial by the first polynomial.
4) Here quotient is 3x2 - 4x + 2, and remainder is 0.
5) As the remainder is 0, (x2 + 3x + 1) is a factor of (3x4 + 5x3 – 7x2 + 2x + 2).

(iii) x3 – 3x + 1,      x5 – 4x3 + x2 + 3x + 1
1) Here first polynomial g(x) = x3 – 3x + 1.
2) Second polynomial p(x) = x5 – 4x3 + x2 + 3x + 1.
3) Divide the second polynomial by the first polynomial.
4) Here quotient is x2 - 1, and the remainder is 2.
5) As the remainder is not 0, (x3 – 3x + 1) is not a factor of (x5 – 4x3 + x2 + 3x + 1).

Q3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and -√(5/3).

Explanation:

1) Let p(x) = 3x4 + 6x3 – 2x2 – 10x – 5 and √(5/3) and -√(5/3) are zeroes of p(x).
2) So, [x - √(5/3)] x [x + √(5/3)] is a factor of 3x4 + 6x3 – 2x2 – 10x – 5.
3) Now we will find other factors of p(x).

Solution:

1) Here [x - √(5/3)] x [x + √(5/3)] is a factor of 3x4 + 6x3 – 2x2 – 10x – 5.
2) [x - √(5/3)] x [x + √(5/3)] = (x2 – 5/3).
3) So, (x2 – 5/3) divides the polynomial 3x4 + 6x3 – 2x2 – 10x – 5.
4) Here (x2 – 5/3) is one factor of 3x4 + 6x3 – 2x2 – 10x – 5 and the other one is 
(3x2 + 6x + 3). i.e. 3(x2 + 2x + 1) = 3(x + 1)(x + 1).
5) Here x = -1, x = -1 are the other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5.
6) All zeroes of 3x4 + 6x3 – 2x2 – 10x – 5 are √(5/3), -√(5/3), - 1, and - 1. 
 
Q4. 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).

Explanation:

1) We know that Dividend = Divisor × Quotient + Remainder
2) If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find
polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x).

Solution:

1) Here 
a) dividend = (x3 – 3x2 + x + 2)
b) divisor = g(x)
c) quotient = (x - 2)
d) remainder = (-2x + 4)
2) Dividend = Divisor × Quotient + Remainder.
(x3 – 3x2 + x + 2) = g(x) x (x - 2) + (-2x + 4)
(x3 – 3x2 + x + 2) - (-2x + 4) = g(x) x (x - 2)
(x3 – 3x2 + x + 2 + 2x - 4) = g(x) x (x - 2)
(x3 – 3x2 + 3x - 2) = g(x) x (x - 2)
q(x) = (x3 – 3x2 + 3x - 2)/(x - 2)
3) Now we will divide x3 – 3x2 + 3x - 2 by x - 2.
4) Therefore g(x) = x2 - x + 1.

Q5. 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

Explanation:

1) We know that the division algorithm is 
Dividend = Divisor × Quotient + Remainder.
2) If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find
polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x).

 Solution:

(i) deg p(x) = deg q(x)
1) Here it is clear that, when divisor g(x) is constant, then degree of dividend p(x)
and quotient q(x) will be same and they will satisfy the division algorithm.
2) Let p(x) be 2x2 - 4x + 6 and g(x) be 2, then q(x) will be x2 - 2x + 3.
3) So here degree of p(x) = degree of q(x) = 2.
4) Now we will prove the division algorithm, 
Dividend = Divisor × Quotient + Remainder
p(x) = g(x) × q(x) + r(x)
LHS = p(x)
= 2x2 - 4x + 6 ----------------------equation 1
RHS = g(x) × q(x) + r(x)
= [(x2 - 2x + 3) x (2)] + 0
= 2x2 - 4x + 6 ----------------------equation 2
5) From equation 1 and equation 2, LHS = RHS.
6) So it satisfies the division algorithm.

(ii) deg q(x) = deg r(x)
1) Let p(x) be 6x3 + 15x2 - x - 17 and g(x) be 3x2 - 2.
2) So, here q(x) is 2x + 5 and r(x) is 3x - 7.
3) So here degree of q(x) = degree of r(x) = 1.
4) Now we will prove the division algorithm, 
Dividend = Divisor × Quotient + Remainder
p(x) = g(x) × q(x) + r(x)
LHS = p(x)
= 6x3 + 15x2 - x - 17 ----------------------equation 1
RHS = g(x) × q(x) + r(x)
= [(3x2 - 2) x (2x + 5)] + 3x - 7
= (6x3 + 15x2 - 4x - 10) + 3x - 7
= 6x3 + 15x2 - x - 17 ----------------------equation 2
5) From equation 1 and equation 2, LHS = RHS.
6) So it satisfies the division algorithm.

(iii) deg r(x) = 0
1) Let p(x) be 12x3 + x2 - 9x + 15 and g(x) be 4x2 + 3x - 1.
2) So, here q(x) is 3x - 2 and r(x) is 13.
3) So here degree of r(x) = 0.
4) Now we will prove the division algorithm, 
Dividend = Divisor × Quotient + Remainder
p(x) = g(x) × q(x) + r(x)
LHS = p(x)
= 12x3 + x2 - 9x + 15 ----------------------equation 1
RHS = g(x) × q(x) + r(x)
= [(4x2 + 3x - 1) x (3x - 2)] + 13
= (12x3 + x2 - 9x + 2) + 13
= 12x3 + x2 - 9x + 15 ----------------------equation 2
5) From equation 1 and equation 2, LHS = RHS.
6) So it satisfies the division algorithm.
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