Haryana State Board HBSE 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.
Haryana Board 10th Class Maths Solutions Chapter 2 Polynomials Exercise 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 :
(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.
Solution :
(i) p(x) = x3 – 3x2 + 5x – 3
g(x) = x2 – 2
Now, we divide p(x) by g(x) as follows :
Hence, quotient q(x) = x – 3 and remainder r(x) = 7x – 9.
(ii) p(x) = x4 – 3x2 + 4x + 5
and g(x) = x2 + 1 – x
We write g(x) in standard form of decreasing power of x.
So, g(x) = x2 – x + 1
Now, we divide p(x) by g(x) as follows :
Hence, quotient q(x) = x2 + x – 3 and remainder r(x) = 8.
(iii) p(x) = x4 – 5x + 6 and g(x) = 2 – x2
We write g(x) in standard form of decreasing power of x.
So, g(x) = – x2 + 2
Now, we divide p(x) by g(x) as follows :
Hence, the quotient q(x) = – x2 – 2 and remainder r(x) = – 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 :
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) xx + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution :
(i) We have,
f(t) = 2t4 + 3t3 – 2t2 – 9t – 12
g(t) = t2 – 3
Now, we divide f(t) by g(t) as follows :
Since, the remainder is zero therefore, the second polynomial is divisible by first polynomial.
Hence, the first polynomial is a factor of the second polynomial.
(ii) We have,
f(x) = 3x4 + 5x3 – 7x2 + 2x + 2 and g(x) = x2 + 3x + 1
Now, we divide f(x) by g(x) as follows :
Since, the remainder is zero therefore, the second polynomial is divisible by first polynomial.
Hence, the first polynomial is a factor of the second polynomial.
(iii) We have,
f(x) = x5 – 4x3 + x2 + 3x + 1
g(x) = x3 – 3x + 1
Now, we divide f(x) by g(x) as follows :
Since remainder = 2 ≠ 0 therefore, second polynomial is not divisible by first polynomial.
Hence, first polynomial is not a factor of second polynomial.
Question 3.
Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)
Solution :
We know that if a is zero of the polynomial then (x – α) is a factor of given polynomial.
Since \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes of the polynomial f(x) = 3x4 + 6x3 – 2x2 – 10x – 5
Hence, (x – \(\sqrt{\frac{5}{3}}\)) (x + \(\sqrt{\frac{5}{3}}\)) = x2 – \(\frac{5}{3}\) is a factor of f(x).
Now, we divide f(x) by x2 – \(\frac{5}{3}\) to obtain the other zeroes as follows :
According to division algorithm of polynomial,
∴ 3x4 + 6x3 – 2x2 – 10x – 5 = (x2 – \(\frac{5}{3}\)) (3x2 + 6x + 3)
= (x2 – \(\frac{5}{3}\)) × 3(x2 + 2x + 1)
= 3(x2 – \(\frac{5}{3}\)) (x + 1)2
For zeroes of the polynomial f(x) = 0
∴ 3 (x2 – \(\frac{5}{3}\)) (x + 1)2 = 0
⇒ (x + \(\sqrt{\frac{5}{3}}\)) (x – \(\sqrt{\frac{5}{3}}\)) (x + 1) (x + 1) = 0
⇒ x = – \(\sqrt{\frac{5}{3}}\), \(\sqrt{\frac{5}{3}}\) , – 1, – 1
Hence, the other zeroes of polynomial are – 1, – 1.
Question 4.
On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).
Solution :
We have,
f(x) = x3 – 3x2 + x + 2
q(x) = x – 2
r(x) = – 2x + 4; g(x) = ?
According to division algorithm of polynomials
f(x) = g(x) × q(x) + r(x)
⇒ f(x) – r(x) = g(x) × q(x)
Hence, g(x) = x2 – x + 1.
Question 5.
Give examples of polynomial p(x), g(x) and Kx), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg Kx)
(iii) deg Kx) = 0.
Solution :
(i) Let p(x) = 4x2 + 4x + 12 and g(x) = 4.
Now, we divide p(x) by g(x) and we get
q(x) = \(\frac{4 x^2+4 x+12}{4}\)
= x2 + x + 3
∴ r(x) = 0.
Hence, p(x) = 4x2 + 4x + 12, g(x) = 4, q(x) = x2 + x + 3, r(x) = 0
∴ deg p(x) = deg q(x)
(ii) Let p(x) = x3 + x2 + x + 1
and g(x) = x2 – 1
Now, we get q(x) and r(x) on dividing p(x) by g(x) as follows :
Hence, p(x) = x3 + x2 + x + 1, g(x) = x2 – 1, r(x) = 2x + 2, q(x) = x + 1
deg q(x) = deg r(x).
(iii) Let p(x) = x3 + 4x2 – x + 7
and g(x) = x2 – 1
Now, we get q(x), r(x) on dividing p(x) by g(x) as follows :
Hence, p(x) = x3 + 4x2 – x + 7, g(x) = x2 – 1, g(x) = x + 4, r(x) = 11.
∴ deg r(x) = 0.