Question

if the polynomial 6x4+8x3-5x2+ax+b is exactly divisible by the polynomial 2x2-5. then find the value of a and b

Answer 1

Hello Mate!

So, according to factor theorum p( x ) which is remainder should be '0' of 2x^2 - 5 is dividing 6x^4 - 8x^3 - 5x^2 + ax + b completely.

So, 2x^2 - 5 = 0
2x^2 = 5
x^2 = 5/2
$x = \sqrt{ \frac{5}{2} } \\ 0 = 6 {( \sqrt{ \frac{5}{2} } )}^{4} - 8 {( \sqrt{ \frac{5}{2} }) }^{3} - 5 {( \sqrt{ \frac{5}{2} } )}^{2} + a( \sqrt{ \frac{5}{2} } ) + b \\ 0 = 6 \times \: \frac{25}{4} - 8 \times \frac{5 \sqrt{5} }{2 \sqrt{2} } - 5 \times \frac{5}{2} + \sqrt{ \frac{5}{2} } a + b \\ 0 = \frac{125}{2} - \frac{20 \sqrt{5} }{ \sqrt{2} } - \frac{25}{2} + \sqrt{ \frac{5}{2} } a + b \\ 0 = \frac{125 - 25}{2} - \sqrt{ \frac{5}{2} } (20 - a) + b \\ 0 = 50 - \sqrt{ \frac{5}{2} } (20 - a) + b \\ - 50 = - \sqrt{ \frac{5}{2} } (20 - a) + b \\ - 50 \times - \sqrt{ \frac{2}{5} } = 20 - a + b \\ \frac{50 \sqrt{2} }{ \sqrt{5} } \times \frac{ \sqrt{5} }{ \sqrt{5} } =20 - a + b\\ 10 \sqrt{10} = 20- a + b - - - > (1)$
Now, lets take x = - root ( 5/2 )

$x = - \sqrt{ \frac{5}{2} } \\ 0 = 6 { ( - \sqrt{ \frac{5}{2} } )}^{4} - 8 {( - \sqrt{ \frac{5}{2} } )}^{3} - 5( { - \sqrt{ \frac{5}{2} } )}^{2} + a( - \sqrt{ \frac{5}{2} }) + b \\ 0 = \frac{125}{2} + 8 \times \frac{5 \sqrt{5} }{2 \sqrt{2} } - \frac{25}{2} - \sqrt{ \frac{5}{2} } a + b \\ 0 = 50 - \sqrt{ \frac{5}{2} } ( - 20 + a) - b \\ 10 \sqrt{10} = - 20 + a + b - - - - - > (2)$

On subtracting two equations we get

$10 \sqrt{10} = 20 - a + b \\ - (10 \sqrt{10} = - 20 + a + b) \\ = > 0 = 40 - 2a \\ - 40 = - 2a \\ a = 20$

Keeping a value in equation 2

$10 \sqrt{10} = - 20 + 20 + b \\10 \sqrt{10} = b$

Hope it helps☺!✌