find the value of k , if y+3 is a factor of 3y^+ky+6
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Answered by
2
y+3=0
y=-3
p(y)=3y²+ky+6
p(-3)=3(-3)²+k(-3)+6
0=3*9-3k+6
0=27+6-3k
3k=33
k=33/3
k=11
The value of k is 11.
y=-3
p(y)=3y²+ky+6
p(-3)=3(-3)²+k(-3)+6
0=3*9-3k+6
0=27+6-3k
3k=33
k=33/3
k=11
The value of k is 11.
Answered by
1
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
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