Find all zeros of the polynomial When appropriate use the Ra

Find all zeros of the polynomial, When appropriate, use the Rational Zeros Theorem, Descartes Rule of Signs, the quadratic formulae, or other factoring techniques. g(x) = x^4 - bx^2 - 8x + 24 x = _______________r(x) = x^3 + 2x^2 - 5x - 6 x = ______________

Solution

g(x) = x^4 - 6x^2 -8x + 24

to find zeros of the function set g(x) = 0 and solve for x

x^4 - 6x^2 -8x + 24 = 0

finding possible rational roots of the equation

+- { 1 , 2, 3,4 , 6 ,8,12,24 }

checking each root we find that actual root occurs at x = 2

hence dividing the equation by x-2

x^4 - 6x^2 -8x + 24 / (x-2 ) = x^3+2x^2 -2x-12

again possible rational roots of this polynomial are

+- { 1, 2, 3 , 4 , 6 , 12 }

actual root occurs at x = 2

again dividing x^3+2x^2 -2x-12 by x-2

we get x^2+4x +6

finding remaining roots by applying quadratic formula

x = { -b + - sqrt (b^2 - 4ac ) } / 2a

x = -2 + i sqrt 2

x = -2 - i sqrt 2

therefore, 4 zeros are

x = 2

x = 2

x = -2 + i sqrt 2

x = -2 - i sqrt 2

b) p(x) = x^3 + 2x^2 -5x - 6

setting p(x) = 0

x^3 + 2x^2 -5x - 6 = 0

possibl rational roots are

+- { 1 , 2 , 3 , 6 }

actual roots occur at x = -1

dividing the polynomila by x+1 we get

x^3 + 2x^2 -5x - 6 / (x+1 )

x^2 + x - 6

factoring the quadratic equation to find other zeros

x^2 + x - 6 = (x+3)(x-2)

hence other zeros are

x+ 3= 0

x = -3

x-2 = 0

x = 2

3 zeros are

x = -1

x = 2

x = -3