What are the zeros of -8x^3-7x^2+x

I am not sure how to find the zeros for this problem can anyone explain how to approach this problem. Thankyou for your time.

-8x³ - 7x² + x = 0
-x(8x² + 7x - 1) = 0 {factored -x out}
-x(8x - 1)(x + 1) = 0 {factored into two binomials}
-x = 0 or 8x - 1 = 0 or x + 1 = 0 {set each factor equal to 0}
x = 0 or 8x = 1 or x = -1 {divided by -1 in 1st, added 1 in 2nd, and subtracted 1 in 3rd}
x = 0 or x = 1/8 or x = -1 {divided 2nd equation by 8}

Factor out the x term that's common to all three parts of the expression:

x(-8x^2 - 7x + 1)

One zero is x = 0.

next factor the (-8x^2 - 7x + 1)

(-8x + 1)(x + 1)

-8x + 1 will be zero when x = 1/8

x + 1 will be zero when x = -1


The three zeroes are -1, 0, and 1/8.

-8x³ - 7x² + x = 0
-x(8x² + 7x - 1) = 0 {factored -x out}
-x(8x - 1)(x + 1) = 0 {factored into two binomials}
-x = 0 or 8x - 1 = 0 or x + 1 = 0 {set each factor equal to 0}
x = 0 or 8x = 1 or x = -1 {divided by -1 in 1st, added 1 in 2nd, and subtracted 1 in 3rd}
x = 0 or x = 1/8 or x = -1 {divided 2nd equation by 8}
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