Question

for what value of a 2xcube+axsquare+11x+a+3 is exactly divided by (2x-1)?

Answer 1

Hi friend!!

Given polynomial f(x)= 2x³+ax²+11x+a+3

If f(x) is exactly divided by 2x-1.Then f(1/2)=0

→2(1/2)³+a(1/2)²+11(1/2)+a+3=0

→(1/4)+(a/4)+(11/2)+a+3=0

5a/4=-(1+22+12)/4

5a=-35

a=-7

I hope this will help you ;)

Answer 2

Given that (2x - 1) is the factor, 

Then,  2x - 1 = 0 
                2x  = 1
                  x = 1/2 

------------------------------------------------

2x³ + ax² + 11x + a + 3 = 0 

2(1/2)³ + a(1/2)² + 11(1/2) + a +  3 = 0 

2(1/8) + a/4  + 11/2 + a + 3  = 0 

1/4 + a/4 + 11/2 + a + 3 = 0 

a/4 + a  = - 3 - 11/2 - 1/4 

(a + 4a)/4  = ( -12 - 22 - 1)/4 

5a   = -12 - 22 -1

5a   = -35

a = -35/5 

a = -7 

i hope this will help you

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