Question

factor to find all x-intercepts of the function.\\(f(x) = -2x^5 - x^4 + 10x^3 + 5x^2\\)\
answer attempt 2 out of 2\
\\(\oplus\\) additional solution \\(\ominus\\) no solution\
\\(x = \square\\)\
\\(\sqrt{}\\) \\(\pm\\) \\(\text{submit answer}\\)

Explanation:

Step1: Factor out the GCF

First, factor out the greatest common factor (GCF) from the polynomial \( f(x) = -2x^5 - x^4 + 10x^3 + 5x^2 \). The GCF of the terms is \( -x^2 \) (we can also factor out \( x^2 \) first, but factoring out \( -x^2 \) might make the quadratic easier to factor later; alternatively, factor out \( x^2 \) first: \( f(x)=x^2(-2x^3 - x^2 + 10x + 5) \)). Wait, let's group the terms in the cubic part. Let's rewrite the polynomial as \( f(x)= -2x^5 - x^4 + 10x^3 + 5x^2 = -x^4(2x + 1) + 5x^2(2x + 1) \). Wait, actually, let's factor by grouping. Group the first two terms and the last two terms:

\( f(x)= (-2x^5 - x^4)+(10x^3 + 5x^2) \)

Factor out \( -x^4 \) from the first group and \( 5x^2 \) from the second group:

\( f(x)= -x^4(2x + 1) + 5x^2(2x + 1) \)

Now, factor out the common binomial factor \( (2x + 1) \):

\( f(x)= (2x + 1)(-x^4 + 5x^2) \)

Factor out \( -x^2 \) from the second factor:

\( f(x)= (2x + 1)(-x^2)(x^2 - 5) \)

Or, we can also factor out \( x^2 \) initially:

\( f(x)=x^2(-2x^3 - x^2 + 10x + 5) \)

Now, factor the cubic polynomial \( -2x^3 - x^2 + 10x + 5 \) by grouping. Group as \( (-2x^3 - x^2)+(10x + 5) \)

Factor out \( -x^2 \) from the first group and \( 5 \) from the second group:

\( -x^2(2x + 1) + 5(2x + 1) \)

Now, factor out \( (2x + 1) \):

\( (2x + 1)(-x^2 + 5) \)

So, putting it back with the \( x^2 \) we factored out initially:

\( f(x)=x^2(2x + 1)(-x^2 + 5) \)

We can rewrite \( -x^2 + 5 \) as \( 5 - x^2 \), which is a difference of squares: \( 5 - x^2 = (\sqrt{5} - x)(\sqrt{5} + x) \), but maybe we made a mistake in factoring. Wait, let's check the initial factoring again. Let's start over.

Given \( f(x) = -2x^5 - x^4 + 10x^3 + 5x^2 \)

Factor out \( -x^2 \) from all terms? Wait, no, the GCF of all terms is \( x^2 \) (since all terms have at least \( x^2 \)). Let's factor out \( x^2 \):

\( f(x) = x^2(-2x^3 - x^2 + 10x + 5) \)

Now, factor the cubic \( -2x^3 - x^2 + 10x + 5 \). Let's factor by grouping. Rearrange the terms: \( -2x^3 + 10x - x^2 + 5 \)

Group as \( (-2x^3 + 10x) + (-x^2 + 5) \)

Factor out \( -2x \) from the first group and \( -1 \) from the second group:

\( -2x(x^2 - 5) -1(x^2 - 5) \)

Now, factor out \( (x^2 - 5) \):

\( (x^2 - 5)(-2x - 1) \)

So, now the polynomial becomes:

\( f(x) = x^2(x^2 - 5)(-2x - 1) \)

We can factor out a negative sign from \( -2x -1 \) to get \( - (2x + 1) \), so:

\( f(x) = -x^2(x^2 - 5)(2x + 1) \)

Or, \( f(x) = x^2(5 - x^2)(2x + 1) \) (since \( - (x^2 - 5) = 5 - x^2 \))

Now, to find the x-intercepts, we set \( f(x) = 0 \):

\( x^2(2x + 1)(5 - x^2) = 0 \)

Set each factor equal to zero:

1. \( x^2 = 0 \) gives \( x = 0 \) (double root)
2. \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \)
3. \( 5 - x^2 = 0 \) gives \( x^2 = 5 \), so \( x = \sqrt{5} \) or \( x = -\sqrt{5} \)

So the x-intercepts are \( x = 0 \), \( x = -\frac{1}{2} \), \( x = \sqrt{5} \), and \( x = -\sqrt{5} \). But let's check the factoring again to make sure.

Wait, let's do the factoring step by step correctly:

Original function: \( f(x) = -2x^5 - x^4 + 10x^3 + 5x^2 \)

Factor out \( -x^2 \) from all terms? Wait, no, the coefficients are -2, -1, 10, 5. The GCF of the coefficients: GCF of 2,1,10,5 is 1, but the variable part is \( x^2 \) (since all terms have \( x^2 \) or higher). So factor out \( x^2 \):

\( f(x) = x^2(-2x^3 - x^2 + 10x + 5) \)

Now, factor the cubic \( -2x^3 - x^2 + 10x + 5 \). Let's use rational root theorem. Possible rational roots are \( \pm1, \pm5, \pm\frac{1}{2}, \pm\frac{5}{2} \). Let's test \( x = -\frac{1}{2} \):

\( -2(-\frac{1}{2})^3 - (-\frac{1}{2})^2 + 10(-\frac{1}{2}) + 5 \)

\( -2(-\frac{1}{8}) - \frac{1}{4} - 5 + 5 \)

\( \frac{1}{4} - \frac{1}{4} - 5 + 5 = 0 \). So \( x = -\frac{1}{2} \) is a root, so \( (2x + 1) \) is a factor (since \( x = -\frac{1}{2} \) implies \( 2x + 1 = 0 \)).

Now, divide the cubic \( -2x^3 - x^2 + 10x + 5 \) by \( (2x + 1) \) using polynomial long division or synthetic division.

Using synthetic division with root \( x = -\frac{1}{2} \):

Coefficients of the cubic: -2 (x³), -1 (x²), 10 (x), 5 (constant)

Set up synthetic division:

-1/2 | -2 -1 10 5

1 0 -5

-------------------

-2 0 10 0

So the quotient is \( -2x^2 + 0x + 10 = -2x^2 + 10 = -2(x^2 - 5) \)

Therefore, the cubic factors as \( (2x + 1)(-2x^2 + 10) = (2x + 1)(-2)(x^2 - 5) = -2(2x + 1)(x^2 - 5) \)

So putting it back into \( f(x) \):

\( f(x) = x^2 \times (-2)(2x + 1)(x^2 - 5) = -2x^2(2x + 1)(x^2 - 5) \)

Setting \( f(x) = 0 \):

\( -2x^2(2x + 1)(x^2 - 5) = 0 \)

Since -2 ≠ 0, we set each factor to zero:

- \( x^2 = 0 \) ⇒ \( x = 0 \)
- \( 2x + 1 = 0 \) ⇒ \( x = -\frac{1}{2} \)
- \( x^2 - 5 = 0 \) ⇒ \( x^2 = 5 \) ⇒ \( x = \sqrt{5} \) or \( x = -\sqrt{5} \)

So the x-intercepts are \( x = 0 \), \( x = -\frac{1}{2} \), \( x = \sqrt{5} \), and \( x = -\sqrt{5} \). But the problem says "all x-intercepts", so we need to list them. However, the input box is for a single x, but maybe the question expects all, but let's check the factoring again. Wait, maybe the initial factoring can be done as:

\( f(x) = -2x^5 - x^4 + 10x^3 + 5x^2 = -x^4(2x + 1) + 5x^2(2x + 1) = (2x + 1)(-x^4 + 5x^2) = (2x + 1)x^2(-x^2 + 5) = x^2(2x + 1)(5 - x^2) \)

Setting to zero:

- \( x^2 = 0 \) ⇒ \( x = 0 \)
- \( 2x + 1 = 0 \) ⇒ \( x = -\frac{1}{2} \)
- \( 5 - x^2 = 0 \) ⇒ \( x^2 = 5 \) ⇒ \( x = \pm\sqrt{5} \)

So the x-intercepts are \( x = 0 \), \( x = -\frac{1}{2} \), \( x = \sqrt{5} \), \( x = -\sqrt{5} \)

But the problem's input box is for a single x, but maybe it's a multiple answer, but the user's input box is a single line. Wait, maybe the question has a typo, or maybe we missed something. Wait, let's check the original function again: \( f(x) = -2x^5 - x^4 + 10x^3 + 5x^2 \). Let's factor out \( x^2 \) first:

\( f(x) = x^2(-2x^3 - x^2 + 10x + 5) \)

Now, factor the cubic \( -2x^3 - x^2 + 10x + 5 \). Let's try to factor by grouping:

\( -2x^3 - x^2 + 10x + 5 = -x^2(2x + 1) + 5(2x + 1) = (2x + 1)(-x^2 + 5) \)

Ah, that's a simpler way! So:

\( f(x) = x^2(2x + 1)(-x^2 + 5) = x^2(2x + 1)(5 - x^2) \)

Now, set \( f(x) = 0 \):

\( x^2(2x + 1)(5 - x^2) = 0 \)

So:

- \( x^2 = 0 \) ⇒ \( x = 0 \)
- \( 2x + 1 = 0 \) ⇒ \( x = -\frac{1}{2} \)
- \( 5 - x^2 = 0 \) ⇒ \( x^2 = 5 \) ⇒ \( x = \sqrt{5} \) or \( x = -\sqrt{5} \)

So the x-intercepts are \( x = 0 \), \( x = -\frac{1}{2} \), \( x = \sqrt{5} \), and \( x = -\sqrt{5} \). Since the problem says "all x-intercepts", but the input box is a single line, maybe it's expecting all of them. However, let's check the problem statement again: "Factor to find all x-intercepts of the function." So we need to list all x-intercepts.

Answer:

The x-intercepts are \( x = 0 \), \( x = -\frac{1}{2} \), \( x = \sqrt{5} \), and \( x = -\sqrt{5} \). If we need to write them in the box (assuming multiple answers are allowed, maybe separated by commas), the answers are \( x = 0 \), \( x = -\frac{1}{2} \), \( x = \sqrt{5} \), \( x = -\sqrt{5} \). But let's confirm the factoring:

\( f(x) = x^2(2x + 1)(5 - x^2) = 0 \)

Solutions:

• \( x^2 = 0 \) ⇒ \( x = 0 \)
• \( 2x + 1 = 0 \) ⇒ \( x = -\frac{1}{2} \)
• \( 5 - x^2 = 0 \) ⇒ \( x = \pm\sqrt{5} \)

So the x-intercepts are \( 0 \), \( -\frac{1}{2} \), \( \sqrt{5} \), and \( -\sqrt{5} \).