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

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(C) Given the cubic equation is $x^3-42x^2+336x-512=0$.We can test for roots. Let $x=2$: $2^3 - 42(2^2) + 336(2) - 512 = 8 - 168 + 672 - 512 = 0$. Thu

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$4$

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(C) Given the cubic equation is $x^3-42x^2+336x-512=0$.We can test for roots. Let $x=2$: $2^3 - 42(2^2) + 336(2) - 512 = 8 - 168 + 672 - 512 = 0$. Thus,$(x-2)$ is a factor.Dividing the polynomial by $(x-2)$,we get $(x-2)(x^2-40x+256)=0$.Solving the quadratic part $x^2-40x+256=0$ using the quadratic formula or factorization:$x = \frac{40 \pm \sqrt{1600 - 1024}}{2} = \frac{40 \pm \sqrt{576}}{2} = \frac{40 \pm 24}{2}$.The roots are $x_1 = \frac{16}{2} = 8$ and $x_2 = \frac{64}{2} = 32$.The roots are $2, 8, 32$.These are in increasing geometric progression with common ratio $r = \frac{8}{2} = 4$ or $4:1$.

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

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