Question

If 6x^4+8x^3-5x^2+ax+b is completely divisible by 2x^2-5 then find the value of a and b.

Answer 1

Answer:

• The value of $a = -20$

• The value of $b=-25$

Step-by-step explanation:

Let

$p(x)=6x^4+8x^3-5x^2+ax+b$

and

$q(x)=2x^2-5$

As $q(x)$ is exactly divisible, so, the remainder is 0.

∴ $p(x)=q(x).g(x)+r(x)$

$6x^4 + 8x^3 - 5x^2 + ax + b = (\sqrt{2} x - \sqrt{5})(\sqrt{2} x + \sqrt{5}) g(x)$

As the R.H.S becomes 0 when $x = \sqrt{\frac{5}{2} }$ and $x = -\sqrt{\frac{5}{2} }$.

And

by substituting $x = \sqrt{\frac{5}{2} }$

$6(\sqrt{\frac{5}{2} } )^4 + 8(\sqrt{\frac{5}{2} } )^3 - 5(\sqrt{\frac{5}{2} } )^2 + a(\sqrt{\frac{5}{2} } ) + b = 0$

$25 + 20( \frac{\sqrt{5}}{\sqrt{2}} ) + a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0...[A]$

by substituting $x = -\sqrt{\frac{5}{2} }$

$25 - 20( \frac{\sqrt{5}}{\sqrt{2}} ) - a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0.....[B]$

Adding Equation [A] and equation [B]

$50 +2b =0$

$b=-25$

Subtracting Equation [A] and equation [B]

$40(\sqrt{\frac{5}{2} } ) + 2a(\sqrt{\frac{5}{2} } ) =0$

$a = -20$

Therefore,

• The value of $a = -20$

• The value of $b=-25$

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