Question

11. Find the value of a and b so that (2x 3 +ax + x+ b ) has (x+2) and (2x-1) as factor

Answer 1

Solution:-

p(x) = 2x³ + ax + x + b

So p(x) is divisible by x + 2

By the factor theorem

(x + 2 ) is factor of p(x)

Then ,

p( - 2) = 0

=> 2x³ + ax + x + b = 0

=> 2 × ( - 2 )³ + a × - 2 - 2 + b = 0

=> 2 × - 8 - 2a - 2 + b = 0

=> - 16 - 2 - 2a + b = 0

=> - 18 - 2a + b = 0

=> 2a - b = - 18 ...........( i ) eq

g(x) = 2x³ + ax + x + b

So g(x) is divisible by ( 2x - 1 )

(2x + 1 ) is factor of g(x)

then

g(1/2) = 0

$2 \times ( \frac{1}{2} ) {}^{3} + a \times \frac{1}{2} + \frac{1}{2} + b = 0$

$\frac{1}{4} {}^{} + \frac{a}{2} + \frac{1}{2} + b = 0$

Take lcm

$\frac{1 + 2a + 2 + 4b}{4} = 0$

$3 + 2a + 4b = 0$

2a + 4b = - 3 .........(ii)

Now we have two equation and two unknown value apply substitution method

2a - b = - 18 .......(i)

2a + 4b = - 3 .........(ii)

Now substitute (i) eq

2a - b = - 18

2a = - 18 + b

$a = \frac{ - 18 + b}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: (iii)eq$

Now put the value of a on (ii)eq

2a + 4b = - 3

$2 \times (\frac{ - 18 + b}{2} ) \: + 4b \: = - 3$

- 18 + b + 4b = - 3

5b = 15

b = 3

Now put the value of b on (iii) eq

$a = \frac{ - 18 + 3}{2}$

$a = \frac{ - 15}{2}$

Answer:- a = -15/2 and b = 3