Question

For what value of a,2y^3+ay^2+11y+a+3 is axactly divisible by 2y-3?

Answer 1

Hi friend,

2y³+ay²+11y+a+3

2y-3 exactly divides the given polynomial.

2y - 3 = 0

2y = 3

y = 3/2

2(3/2)³+a(3/2)²+11(3/2)+a+3 = 0

2(27/8)+a(9/4)+33/2+a+3 = 0

27/4 + 9a/4 + 33/2 +a = -3

(27+9a+66+4a)/4 = -3

93+13a = -12

13a = -12-93

13a = -105

a = -105/13

hope it helps

Answer 2

Hi friend!!!

Given, 2y³+ay²+11y+a+3 is divisible by 2y-3 .since 2y-3 is a factor, y=3/2 is the zero of the polynomial 2y³+ay²+11y+a+3

so f(3/2)=0

→2(27/8)+a(9/4)+11(3/2)+a+3=0

→27/4+9a/4+33/2+a+3=0

→27+9a+66+12+4a=0

→105+13a=0

→13a=-105

a=-105/13

I hope this will help u :)