If the remainder on division of x³ + 2x² + kx +3 by x - 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx - 18
Solution:
Given, p(x) = x³ + 2x² + kx +3
g(x) = x - 3
Remainder, r(x=3) = 21.
We have to find the quotient and the value of k and the zeros of the cubic polynomial.
The division algorithm states that given any polynomial p(x) and any non-zero
polynomial g(x), there are polynomials q(x) and r(x) such that
p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).
To find the value of k, put x = 3 in p(x),
p(3) = (3)³ + 2(3)² + k(3) +3 = 21
27 + 18 + 3k + 3 = 21
48 + 3k = 21
3k = -27
k = -27/3
k = -9
Therefore, the cubic polynomial is x³ + 2x² - 9x + 3.
Using long division to find the quotient,
The quotient is x² + 5x + 6.
On factoring,
x² + 5x + 6 = 0
x² +3x + 2x + 6 = 0
x(x + 3) + 2(x + 3) = 0
(x + 2)(x + 3) = 0
Now, x + 2 = 0
x = -2
Also, x + 3 = 0
x = -3
Therefore, the zeros of the polynomial are 3, -2 and -3
✦ Try This: If the remainder on division of x³ + 2x² + kx +3 by x + 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx - 18
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 2
NCERT Exemplar Class 10 Maths Exercise 2.4 Solved Problem 2
If the remainder on division of x³ + 2x² + kx +3 by x - 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx - 18
Summary:
If the remainder on division of x³ + 2x² + kx +3 by x - 3 is 21,the quotient is x² + 5x + 6 and the value of k is -9. Hence, the zeroes of the cubic polynomial x³ + 2x² + kx - 18 are 3, -3 and -2
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