Solve the equation by factoring: x^(3)-7x^(2)-8x=0 Answer: x=

Factor GCF: First, we need to factor out the greatest common factor (GCF) from the equation x^3 - 7x^2 - 8x = 0. The GCF is x. So we factor x out of each term: x(x^2 - 7x - 8) = 0.Factor Quadratic Equation: Next, we need to factor the quadratic equation x^2 - 7x - 8. We look for two numbers that multiply to -8 and add up to -7. These numbers are -8 and +1. So we can write the quadratic as (x - 8)(x + 1).Find Solutions: Factor 1: Now we have the factored form of the original equation: x(x - 8)(x + 1) = 0. To find the solutions, we set each factor equal to zero and solve for x.Find Solutions: Factor 2: Setting the first factor equal to zero gives us x = 0.Find Solutions: Factor 3: Setting the second factor equal to zero gives us x - 8 = 0, which means x = 8.Setting the third factor equal to zero gives us x + 1 = 0, which means x = -1.

Home

Math Problems

Precalculus

Operations with rational exponents

Full solution

Q. Solve the equation by factoring:

\newline

x^{3}-7 x^{2}-8 x=0

\newline

Answer:

x=

Factor GCF: First, we need to factor out the greatest common factor (GCF) from the equation $x^3 - 7x^2 - 8x = 0$ . The GCF is $x$ . So we factor $x$ out of each term: $x(x^2 - 7x - 8) = 0$ .
Factor Quadratic Equation: Next, we need to factor the quadratic equation $x^2 - 7x - 8$ . We look for two numbers that multiply to $-8$ and add up to $-7$ . These numbers are $-8$ and $+1$ . So we can write the quadratic as $(x - 8)(x + 1)$ .
Find Solutions: Factor $1$ : Now we have the factored form of the original equation: $x(x - 8)(x + 1) = 0$ . To find the solutions, we set each factor equal to zero and solve for $x$ .
Find Solutions: Factor $2$ : Setting the first factor equal to zero gives us $x = 0$ .
Find Solutions: Factor $3$ : Setting the second factor equal to zero gives us $x - 8 = 0$ , which means $x = 8$ .
Find Solutions: Factor $3$ : Setting the second factor equal to zero gives us $x - 8 = 0$ , which means $x = 8$ .Setting the third factor equal to zero gives us $x + 1 = 0$ , which means $x = -1$ .