Question

Factor each and find all the zeros. Show steps to solve.

a) F(x) = x⁴ + 125x
b) F(x) = x⁴ - 8x

Answer 1

a) the factored form of
`\( F(x) \) is \( x(x + 5)(x^2 - 5x + 25) \)`.

b) the factored form of
`\( F(x) \) is \( x(x - 2)(x^2 + 2x + 4) \)`.

Explanation:

Let's factor each expression and find all the zeros:

a)
`\( F(x) = x^4 + 125x \)`

To factor
`\( x^4 + 125x \)`, we can first factor out the common factor, which is x:

`\[ F(x) = x(x^3 + 125) \]`

Now, notice that
`\( x^3 + 125 \)` is the sum of cubes
`(\( a^3 + b^3 \))`, where a = x and b = 5. The sum of cubes formula is
`\( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)`. Applying this to
`\( x^3 + 125 \)`:

`\[ F(x) = x(x + 5)(x^2 - 5x + 25) \]`

So, the factored form of
`\( F(x) \) is \( x(x + 5)(x^2 - 5x + 25) \).`

Now, set each factor equal to zero and solve for x to find the zeros:

1. x = 0

2.
`\( x + 5 = 0 \) \(\Rightarrow\) \( x = -5 \)`

3.
`\( x^2 - 5x + 25 = 0 \)` is a quadratic equation, and you can use the quadratic formula or complete the square to find the remaining zeros. In this case, the quadratic does not have real roots, so the zeros are complex conjugates.

b)
`\( F(x) = x^4 - 8x \)`

To factor
`\( x^4 - 8x \)`, we can first factor out the common factor, which is x:

`\[ F(x) = x(x^3 - 8) \]`

Now, notice that
`\( x^3 - 8 \)` is the difference of cubes
`(\( a^3 - b^3 \))`, where a = x and b = 2. The difference of cubes formula is
`\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)`. Applying this to
`\( x^3 - 8 \)`:

`\[ F(x) = x(x - 2)(x^2 + 2x + 4) \]`

So, the factored form of
`\( F(x) \)` is
`\( x(x - 2)(x^2 + 2x + 4) \).`

Now, set each factor equal to zero and solve for x to find the zeros:

1. x = 0

2.
`\( x - 2 = 0 \) \(\Rightarrow\) \( x = 2 \)`

3.
`\( x^2 + 2x + 4 = 0 \)` is a quadratic equation, and you can use the quadratic formula or complete the square to find the remaining zeros. In this case, the quadratic does not have real roots, so the zeros are complex conjugates.