Obtain other zeros of x4-7x3 +17x2-17x+6 , if two of its zeros are 1 and 2.
Answers
Answer:
The zeroes of the equation \bold{x^{4}-7 x^{3}+17 x^{2}-17 x+6}x
4
−7x
3
+17x
2
−17x+6 are 1 and 2
Step-by-step explanation:
To find the roots of x^{4}-7 x^{3}+17 x^{2}-17 x+6x
4
−7x
3
+17x
2
−17x+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((x−3)(x−1)(x±c)(x±d))=x
4
−7x
3
+17x
2
−17x+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((x−3)(x−1)(x±c)(x±d))=x
4
−7x
3
+17x
2
−17x+6
Hence dividing \frac{x^{4}-7 x^{3}+17 x^{2}-17 x+6}{(x-3)(x-1)}
(x−3)(x−1)
x
4
−7x
3
+17x
2
−17x+6
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}
(x−3)(x−1)
x
4
−7x
3
+17x
2
−17x+6
=
x
2
−4x+3
x
4
−7x
3
+17x
2
−17x+6
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)
(x−3)(x−1)
x
4
−7x
3
+17x
2
−17x+6
=(x−2)(x−1)
Hence, the other two factors are 1 and 2.
I hope its help for u..
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