Question

obtain other zeros of x4-7x3 +17x2-17x+6 , if two of its zeros are 3 and 1.
NOTE: x means not multiply.

Answer 1

I hope it's helps you

Answer 2

Answer:

The zeroes of the equation $\bold{x^{4}-7 x^{3}+17 x^{2}-17 x+6}$ are 1 and 2

Step-by-step explanation:

To find the roots of $x^{4}-7 x^{3}+17 x^{2}-17 x+6$, we need to assume that every power 4 quadratic equation has 4 roots or 4 zeroes.

Meaning if we take 4 zeroes or 4 roots we take zeroes as   α=3,β=1,γ=c,δ=d, therefore let us take the unknown root as  γ=c,δ=d.

So the above question can be written as

$((x-3)(x-1)(x \pm c)(x \pm d)) =x^{4}-7 x^{3}+17 x^{2}-17 x+6$

That means the solution can be formed as x$((x-3)(x-1)(x \pm c)(x \pm d)) =x^{4}-7 x^{3}+17 x^{2}-17 x+6$

Hence dividing $\frac{x^{4}-7 x^{3}+17 x^{2}-17 x+6}{(x-3)(x-1)}$ we get

$\frac{x^{4}-7 x^{3}+17 x^{2}-17 x+6}{(x-3)(x-1)}=\frac{x^{4}-7 x^{3}+17 x^{2}-17 x+6}{x^{2}-4 x+3}$

On cross multiplication and simplifying,  

The other zeroes formed by division of $\frac{x^{4}-7 x^{3}+17 x^{2}-17 x+6}{(x-3)(x-1)}=(x-2)(x-1)$

Hence, the other two factors are 1 and 2.