If x-3 is a factor of polynomial p(x)=x^3-2x^2+kx+6 then find value of k
Answers
Answer: p(x) = x^3-2x^2+kx +6
g(x) = x-3
To determine the zero of g(x)
g(x) = 0
x-3=0
x= -3
By remainder theorem
p(x) = -3
P(-3) = (-3)^3-2×(-3)^2+k×(-3)+ 6
= -27-2×9-3k+6
= -27-18-3k +6
= -45-3k+6
By factor theorem
P(-3) = 0
= -45-3k+6=0
=-39-3k=0
= -3k=39
= k= 39/-3
= -13
Step-by-step explanation:
Answer:
k = -5
Step-by-step explanation:
ATQ,
(x - 3) is a factor of the polynomial p(x)= x³ - 2x² + kx + 6
⇒ x - 3 = 0
⇒ x = 3
When we substitute this value of "x" in the given polynomial, it'll be equal to 0.
⇒ p(x) = 0
⇒ 0 = x³ - 2x² + kx + 6
⇒ 0 = (3)³ - 2(3)² + k(3) + 6
⇒ 0 = 27 - 2(9) + 3k + 6
⇒ 0 = 27 - 18 + 6 + 3k
⇒ 0 = 9 + 6 + 3k
⇒ 0 = 15 + 3k
⇒ 3k = - 15
⇒ k = -15/3
⇒ k = -5
∴ The value of k is -5.
Cross checking the answer by substituing the value of k:
⇒ p(x) = 0
⇒ 0 = x³ - 2x² + kx + 6
⇒ 0 = (3)³ - 2(3)² + (-5)(3) + 6
⇒ 0 = 27 - 2(9) + -15 + 6
⇒ 0 = 27 - 18 - 15 + 6
⇒ 0 = 0
⇒ LHS = RHS
Hence verified.