If the roots of x^3-42 x^2+336 x-512=0 are in | vedclass

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(C) Given,the cubic equation is $x^3-42 x^2+336 x-512=0$. We can factorize the equation by testing for roots. For $x=2$,$8 - 42(4) + 336(2) - 512 = 8

Vedclass · Accepted Answer

$4$

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(C) Given,the cubic equation is $x^3-42 x^2+336 x-512=0$. We can factorize the equation by testing for roots. For $x=2$,$8 - 42(4) + 336(2) - 512 = 8 - 168 + 672 - 512 = 0$. Thus,$(x-2)$ is a factor. Dividing the cubic by $(x-2)$,we get $x^2(x-2) - 40x(x-2) + 256(x-2) = 0$. This simplifies to $(x-2)(x^2-40x+256) = 0$. Factoring the quadratic part: $x^2-32x-8x+256 = (x-32)(x-8) = 0$. So,the roots are $x = 2, 8, 32$. These roots are in increasing geometric progression with the first term $a = 2$ and common ratio $r = \frac{8}{2} = 4$. Therefore,the common ratio is $4: 1$.

If the roots of $x^3-42 x^2+336 x-512=0$ are in increasing geometric progression,then its common ratio is (in $: 1$)

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