SOLUTION: What is the expression of lowest degree by which (X^2+3X-10)(X^2-X-2) must be multiplied to give a perfect square? Find the square root of (6X^2+7X-3)(3X^2+2X-1)

Question 18339: What is the expression of lowest degree by which (X^2+3X-10)(X^2-X-2) must be multiplied to give a perfect square?
Find the square root of (6X^2+7X-3)(3X^2+2X-1)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
What is the expression of lowest degree by which (X^2+3X-10)(X^2-X-2) must be multiplied to give a perfect square?
(X^2+3X-10)(X^2-X-2)=(x^2+5x-2x-10)(x^2-2x+x-2)=[x(x+5)-2(x+5)][x(x-2)+1(x-2)]
=(x+5)(x-2)(x-2)(x+1)=(x+5)(x+1)(x-2)^2..for this to be perfect square ,we have to multiply with (x+5)(x+1)=x^2+6x+5
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Find the square root of (6X^2+7X-3)(3X^2+2X-1)
(6X^2+7X-3)(3X^2+2X-1)=(6x^2+9x-2x-3)(3x^2+3x-x-1)
=[3x(2x+3)-1(2x+3)][3x(x+1)-1(x+1)]
=(3x-1)(2x+3)(3x-1)(x+1)=[(3x-1)^2](x+1)(2x+3)
hence square root of (6X^2+7X-3)(3X^2+2X-1)
=(3x-1)squarerootof(x+1)(2x+3)...further simplification is not possible .if thereis some mistake in the problem check and correct to find if we get a perfect square root.
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