Answer:
Step-by-step explanation:
Let
[tex]p(x)=6x^4+8x^3-5x^2+ax+b[/tex]
and
[tex]q(x)=2x^2-5[/tex]
As [tex]q(x)[/tex] is exactly divisible, so, the remainder is 0.
∴ [tex]p(x)=q(x).g(x)+r(x)[/tex]
[tex]6x^4 + 8x^3 - 5x^2 + ax + b = (\sqrt{2} x - \sqrt{5})(\sqrt{2} x + \sqrt{5}) g(x)[/tex]
As the R.H.S becomes 0 when [tex]x = \sqrt{\frac{5}{2} }[/tex] and [tex]x = -\sqrt{\frac{5}{2} }[/tex].
And
by substituting [tex]x = \sqrt{\frac{5}{2} }[/tex]
[tex]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[/tex]
[tex]25 + 20( \frac{\sqrt{5}}{\sqrt{2}} ) + a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0...[A][/tex]
by substituting [tex]x = -\sqrt{\frac{5}{2} }[/tex]
[tex]25 - 20( \frac{\sqrt{5}}{\sqrt{2}} ) - a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0.....[B][/tex]
Adding Equation [A] and equation [B]
[tex]50 +2b =0[/tex]
[tex]b=-25[/tex]
Subtracting Equation [A] and equation [B]
[tex]40(\sqrt{\frac{5}{2} } ) + 2a(\sqrt{\frac{5}{2} } ) =0[/tex]
[tex]a = -20[/tex]
Therefore,
Keywords: equation, solution
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