Question

find the value of k , if y+3 is a factor of 3ysq + ky + 6

Answer 1

Hey! The answer goes like this :-

Zero  of y+3
Put, y+3=0
       y= -3

Using Factor Theorem :-

P(y)= 3y sq.+ky+6
P(-3)=3(-3)sq. +k(-3)+6
        = 27-3k+6
        = 33-3k

Equating 33-3k to 0. We get :-

33-3k=0
-3k= -33
k= -33/3
k= -11

Hence the value of k is -11.

I hope it will help u.

Answer 2

Solution :

If y+3 is a factor of the given polynomial 3y^2 + my + 6 = 0

→ y + 3 = 0

→ y = -3

Now , substitute the value of y in the given polynomial,

→ 3y² + ky + 6

→ 3(-3)² + (-3)k + 6 = 0

We equate the equation to zero, because the substitution of the factor always gives the value 0.

When solving the equation ,

→ 27 - 3k + 6 = 0

→ 3k = 33

→ K = 33/3

→ K = 11