If the remainder on dividing x³ + 2x² + Rx + 3 by x–3 is 21, find the quotient and the value of R . Hence find the zeroes of the cubic polynomial 6x³ + √2x² – 10x –4√2
Answers
Answer:
R = -9
x² +5x + 6
√2 , -1/√2 , -2√2/3
Step-by-step explanation:
If the remainder on dividing x³ + 2x² + Rx + 3 by x–3 is 21, find the quotient and the value of R
f(x) = x³ + 2x² + Rx + 3
f(3) = 21
=> (3)³ + 2(3)² + 3R + 3 = 21
=> 27 + 18 + 3R + 3 = 21
=> R = -9
f(x) = x³ + 2x² -9x + 3
as Remainder = 21 when divided by x-3
=> f(x) - 21 is divisible by x - 3
x³ + 2x² -9x + 3 - 21 = x³ + 2x² -9x - 18
f(x) = g(x)(qx)
g(x) = x-3 & q(x) = ax² + bx + c
(ax² + bx + c)(x-3)
= ax³ -3ax² +bx² -3bx +cx -3c
= ax³ + (b-3a)x² + (c-3b)x - 3c
Comparing with
x³ + 2x² -9x - 18
a = 1
b-3a = 2 => b - 3 = 2 => b = 5
-3c = -18 => c = 6
or c-3b = -9 => c -15 =-9 => c = 6
q(x) = x² +5x + 6
6x³ + √2x² – 10x –4√2
x = √2 is factor
as f(√2) = 12√2 + 2√2 -10√2 -4√2 = 0
g(x) = (x - √2)
f(x) = g(x)(qx)
q(x) = ax² + bx + c
(ax² + bx + c)(x - √2)
= ax³ + (b - a√2)x² + (c -b√2)x - c√2
Equating with
6x³ + √2x² – 10x –4√2
a = 6
b - a√2 = √2 => b - 6√2 = √2 => b = 7√2
c√2 = 4√2 => c = 4
or c -b√2 = -10 => c - 14 = -10 => c = 4
q(x) = 6x² + 7√2x + 4
= 6x² + 3√2x + 4√2x +4
= 3√2x(√2x + 1) + 4(√2x + 1)
= (3√2x + 4)(√2x + 1)
√2 , -1/√2 , -2√2/3