Question

If (x + 5) is a factor of

2x³ + kx² - 11x - 30, then value of k be​

Answer 1

Answer:

k = 9

Explanation:

Given polynomial,

⇒ 2x³ + kx² - 11x - 30

Whose one of the factor is (x + 5)

By factor theorem,

⇒ (x + 5) = 0

⇒ x + 5 = 0

⇒ x = -5

Substituting the value of ‘x’ in the polynomial,

⇒ 2x³ + kx² - 11x - 30 = 0

⇒ 2(-5)³ + k(-5)² - 11(-5) - 30 = 0

⇒ 2(-125) + 25k + 55 - 30 = 0

⇒ - 250 + 25k + 55 - 30 = 0

⇒ - 280 + 55 + 25k = 0

⇒ - 225 + 25k = 0

⇒ 25k = 225

⇒ k = 225/25

⇒ k = 9

∴ The value of ‘k’ is 9

_____________________

Verification:

⇒ 2x³ + kx - 11x - 30 = 0

Let's substitute x and k :-

⇒ 2(-5)³ + 9(-5)² - 11(-5) - 30 = 0

⇒ - 250 + 225 + 55 - 30 = 0

⇒ - 280 + 280 = 0

⇒ 0 = 0

Since, L. H. S. = R. H. S.

Hence, verified!!

Answer 2

Answer:

• P(x) = 2x³+kx²-11x-30

And (x+5) is a factor of the above polynomial.

There fore, x+5=0

Or, x=-5

• P(-5)=2×(-5)³+k(5)²-11(-5)-30

= 2×(-125) + 25k+55-30

=-250+25+25k

=-225 +25k

Now, -225+25k=0

or, 25k=225

or, k=9.