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
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2q(x) = x - 3, r(x) = 7x - 9 Verification : p(x) = g(x) x q(x) + r(x) = (x² – 2)(x - 3) + 7x - 9 = x³ – 3x² - 2x + 6 + 7x – 9 = x³ – 3x² + 5x – 3
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – xq(x) = x² + x - 3, r(x) = 8 Verification : p(x) = g(x) x q(x) + r(x) = (x² + 1 – x )(x² + x - 3) + 8 = x⁴ + x³ – 3x² + x² + x - 3 - x³ - x² + 3x + 8 = x⁴ – 3x² + 4x + 5
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²q(x) = -2 – x², r(x) = -5x + 10 Verification : p(x) = g(x) x q(x) + r(x) = (2 – x²)(-2 – x²) + (-5x + 10) = -(4 - x⁴) - 5x + 10 = x⁴ – 5x + 6
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) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12The remainder is 0. So, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.
(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2The remainder is 0. So, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2 .
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1The remainder is not 0. So, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1 .
Question-3 :- Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √5/3 and -√5/3.
Solution :-p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5 Given that two zeroes are √5/3 and -√5/3. So, (x - √5/3)(x + √5/3) = x² - 5/3 [a² - b² = (a + b)(a - b)]q(x) = 3x² + 6x + 3 Now, p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5 = (x - √5/3)(x + √5/3)(3x² + 6x + 3) = (x - √5/3)(x + √5/3) x 3 x (x² + 2x + 1) = (x - √5/3)(x + √5/3) x 3 x (x + 1)² = (x - √5/3)(x + √5/3) x 3 x (x + 1)(x + 1) x + 1 = 0 x = -1 Therefore, all zeroes are √5/3, -√5/3, -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 p(x) = g(x) x q(x) + r(x) x³ – 3x² + x + 2 = g(x) x (x - 2) + (-2x + 4) x³ – 3x² + x + 2 + 2x - 4 = g(x) x (x - 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
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
(i) deg p(x) = deg q(x) We take an example in which deg p(x) = deg q(x) = 2![]()
(ii) deg q(x) = deg r(x) We take an example in which deg p(x) = deg r(x) = 2![]()
(iii) deg r(x) = 0 We take an example in which deg deg r(x) = 0![]()