find the bounds on the real zeros of the following equation in the real number system f(x)=x^4+7x^3+4x^2-x-1 show your work.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the rational zero test to find the bounds of real zeroes.

You should know that any rational zero of your polynomial will be of the form D_(-1)/D_(1) (divisors of the last coefficient over the divisors of the leading coefficient).

D_(-1) = {+-1} and D_(1) = {+-1}

Plugging the value -1 in the polynomial yields:

f(-1) = (-1)^4 + 7(-1)^3 + 4(-1)^2 - (-1) - 1

f(-1) = 1 - 7 + 4 + 1 - 1 = -2 != 0 => -1 is not a zero for f(x).

You need to find the derivative of f(x) to test Descartes' rule of signs.

f'(x) = 4x^3 + 21x^2+ 8x - 1

f(-1/2) = -4/8 + 21/4 - 8/2 - 1 = (-4 + 42 - 32 - 8)/8 = -2/8!=0

f(1/2) = 4/8 + 21/4 + 8/2 - 1 != 0

f(-1/4) = -4/64 + 21/16 - 8/4 - 1 = (-4 + 84 - 128 - 16)/8 != 0

There are no positive real zeroes of derivative.

You need to take a look at the number of variations is sign of f(x).

f(x)=x^4+7x^3+4x^2-x-1 (one variation in sign means one negative real zero of f(x)).

You need to take a look at the number of variations is sign of f(-x).

f(-x)=x^4-7x^3+4x^2+x-1 (three variations in sign mean three  real zeros of f(-x) or one positive real zero).

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