Find all the real zeros of the polynomial.Enter all answers including repetitions. P(x) = 2x^4 − x^3 − 10x^2 + 3x + 10

Answer 1

p={+-1, +-2, +-5, +-10}

q={+-1, +-2}

p/q = {+-1, +-2, +-5, +-10,+-5/2}

P(-1) = 2 - (-1) - 10 - 3 + 10

= 3 - 10 -3 + 10

= 0

P(2) = 32 - 8 - 40 + 6 + 10

= 24 - 40 + 6 + 10

= -16 + 16

= 0

So -1 and 2 are rational, in fact integer, solutions

The factors are (x+1)(x-2) = x^2 - x- 2

Dividing

2x^2 + x - 5

____________________________

x^2 - x - 2 | 2x^4 - x^3 - 10x^2 + 3x + 10

2x^4 - 2x^3 - 4x^2

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x^3 - 6x^2 + 3x

x^3 - x^2 - 2x

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-5x^2 + 5x + 10

Optionally you can do synthetic division twice:

-1 | 2 -1 - 10 3 10

-2 3 7 -10

__________________________________

2 -3 -7 10 0

2 | 2 -3 -7 10

4 2 -10

____________________

2 1 -5 0

2x^2 +x - 5 =0

the other two solutions per quadratic formula are:

[-1 +or- sqrt ( 1 - 4*2*-5)] / 4 =

[-1 +or- sqrt( 41)]/4 =

which are APPROXIMATELY x=1.350781593582121716220544186555...

and x=-1.8507810593582121716220544186555

The complete solution set is {-1,2, (-1 +or- sqrt(41))/4 }

Answer 2

By trial and error, -1 and 2 are roots. Divide P(x) synthetically by x + 1 and then by x - 2:

-1l 2 -1 -10 3 10

__ -2 3__7__-10

2 -3 -7 10 0

So, P(x) = (x + 1)(2x³ - 3x² - 7x + 10)

2l 2 -3 -7 10

_____4___2___-10

2 1 -5 0

P(x) = (x + 1)(x - 2)(2x² + x - 5)

By the quadratic formula, the remaining roots are x = [-1 ± √41] / 4