Question

Find the value of 'a' for which 2x3+ax2+11x+a+3 is exactly divisible by 2x-1
Please show the whole solution also because I tried many times to do it but I can't able to do so please show the whole solution

Answer 1

Dear Student,

Answer: a = -7

Solution:

from polynomial 2x-1

put the value of x,in cubic polynomial,and solve for a.

$2x - 1 = 0 \\ \\ 2x = 1 \\ \\ x = \frac{1}{2}$
if given polynomial is exactly divisible by 2x-1, than on putting x = 1/2, it becomes zero.
$2 {x}^{3} + a {x}^{2} + 11x + a + 3 \\ \\ 2( { \frac{1}{2}) }^{3} + a( { \frac{1}{2} )}^{2} + 11( \frac{1}{2} ) + a + 3 = 0 \\ \\ 2 \times \frac{1}{8} + a \times \frac{1}{4} + \frac{11}{2} + a + 3 = 0 \\ \frac{1}{4} + \frac{a}{4} + \frac{11}{2} + a + 3 = 0 \\ \frac{a}{4} + a = - 3 - \frac{11}{2} - \frac{1}{4} \\ \\ \frac{a + 4a}{4} = \frac{ - 12 - 22 - 1}{4} \\ \\ 5a = - 35 \\ \\ a = \frac{ - 35}{5} \\ \\ a = - 7$
The value of a = -7.

Hope it helps you.

Answer 2

GIVEN : Let Polynomial f(x)=2x³+ax²+11x+a+3 …….(1)

Polynomial is exactly divisible by 2x-1 if the remainder is 0.

2x -1= 0

2x = 1

x = ½

By remainder theorem, when f(x) is divided by (2x-1) then remainder = f(½)

On Putting x= ½ in eq 1

f(1/2)=2(1/2)³+a(1/2)²+11(1/2)+a+3

0 =2(⅛)+a/4+11/2+a+3

0 = ¼+a/4+11/2+a+3

0 = ¼ + 11/2 +3 +a/4 +a

0 = (1+22+12)/4 +(a + 4a)/4

0= 35/4 + 5a/4

-35/4 = 5a/4

-35 = 5a

a = -35/5

a = -7

Hence, the value of a is -7

HOPE THIS ANSWER WILL HELP YOU…