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solve by finding all roots: x^4-5x^3+7x^2-5x+6=0

0 votes

finding all roots.

asked Feb 24, 2014 in ALGEBRA 2 by homeworkhelp Mentor

3 Answers

0 votes

Given x^4-5x^3+7x^2-5x+6 = 0

By synthatic division

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x^4-5x^3+7x^2-5x+6 = (x-2)(x-3)(x^2+1)

Now solve x^2+1

Compare it to quadratic form ax^2+bx+c .

a = 1, b = 0, c = 1.

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Roots of x^4-5x^3+7x^2-5x+6 is x = 2,3,i,-i.

answered Feb 24, 2014 by dozey Mentor
0 votes

x4 - 5x+ 7x– 5x + 6 = 0

Rational Root Theorem, if a rational number in simplest form p /q is a root of the polynomial equation anxn + an  1xn – 1 + ... + a1x + a0 = 0, then p is a factor of a0 and q is a factor if an.

If p /q is a rational zero, then p  is a factor of 6 and q  is a factor of 1.

The possible values of p  are   ± 1,   ± 2, ± 3.

The possible values for q  are ± 1

By the Rational Roots Theorem, the only possible rational roots are, p / = ± 1,   ± 2,   ±3

Make a table for the synthetic division and test possible real zeros.

p /q

1

-5

7

-5

6

1

1

-4

3

-1

7

-1

1

6

14

27

60

-2

1

-7

21

-47

100

2

1

-3

1

-3

0

Since f (2) = 0,   x = 2 is a zero. The depressed polynomial is  x3 - 3x2 + x - 3 = 0.

answered May 15, 2014 by david Expert
0 votes

Continuous...

Now x3 - 3x2 + x - 3 = 0.

If p/q is a rational zero, then p  is a factor of 3 and q  is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p /q  = ± 1,   ± 3.

Make a table for the synthetic division and test possible real zeros.

p /q

1

-3

1

-3

1

1

-2

-1

-4

-1

1

-4

5

-8

3

1

0

1

0

Since f (3) =  0, x = 3 is a zero. The depressed polynomial is  x + 1 = 0

image

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Roots of the polynomial  at x = 3 and x = 2 and two imaginary roots at x = i and x = -i.

answered May 15, 2014 by david Expert

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