Find Rational Zeros

Find (if possible) the rational zeros of

f(x) = x^3 + 2x^2 + 6x − 4

The leading coefficient is 1, so the possible rational zeros are the factors of the constant term.

Possible rational zeros: -1, 1, -2, 2, -4, 4

f(-1) = (-1)^3 + 2(-1)^2 + 6(-1) - 4

f(-1) = -1 + 2 - 6 - 4

f(-1) = -9

f(1) = (1)^3 + 2(1)^2 + 6(1) - 4

f(1) = 1 + 2 + 6 - 4

f(1) = 9 - 4

f(1) = 5

f(-2) = (-2)^3 + 2(-2)^2 + 6(-2) - 4

f(-2) = -8 + 8 - 12 - 4

f(-2) = -12 - 4

f(-2) = -16

f(2) = (2)^3 + 2(2)^2 + 6(2) - 4

f(2) = 8 + 8 + 12 - 4

f(2) = 28 - 4

f(2) = -4

f(-4) = (-4)^3 + 2(-4)^2 + 6(-4) - 4

f(-4) = -64 + 32 - 24 - 4

f(-4) = - 60

f(4) = (4)^3 + 2(4)^2 + 6(4) - 4

f(4) = 64 + 32 + 24 - 20

f(4) = 100

So, the given polynomial has no rational zeros.

 

perfect

 

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