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.
Answers
Answer:
Value of k = -9 and Zeroes of given cubic polynomial = 3 , -2 , -3
Step-by-step explanation:
Given: Dividend = x³ + 2x² + kx + 3 and Divisor = x - 3
Remainder = 21
Cubic Polynomial, p(x) = x³ + 2x² + kx - 18
To find: value of k and Zeroes of p(x)
To find value of k We use Remainder Theorem which states that if a polynomial q(x) is divided by x - a then q(a) is remainder of that division.
let q(x) = x³ + 2x² + kx + 3
⇒ Remainder = q(3)
21 = 3³ + 2 × 3² + k × 3 + 3
27 + 18 + 3 +3k = 21
3k = 21 - 48
3k = -27
k = -9
So, Given cubic polynomial, p(x) = x³ + 2x² - 9x - 18
Now to find all zeroes we first we find a zero by hit and trial method
put x = 3
we get,
p(3) = 3³ + 2 × 3² - 9 × 3 + 18 = 27 + 18 -18 -27 = 0
⇒ ( x - 3 ) is a factor of p(x) and 3 is its frst zero.
To find left zero we divide p(x) by x - 3
Division is attached in pic.
we get,
p(x) = ( x - 3 ) ( x² + 5x + 6 )
= ( x - 3 ) ( x² + 2x + 3x + 6 )
= ( x - 3 ) ( x ( x + 2 ) + 3 ( x + 2) )
= ( x - 3 ) ( x + 2 ) ( x + 3)
⇒ Zeroes are 3 , -2 , -3
Therefore, Value of k = -9 and Zeroes of given cubic polynomial = 3 , -2 , -3
