Question

or to find all x-intercepts of the function.
$f(x) = 2x^4 + 9x^3 + 8x^2 + 36x$
answer attempt 1 out of 2

Explanation:

Step1: Factor out the common term

First, we can factor out an \( x \) from each term of the function \( f(x) = 2x^4 + 9x^3 + 8x^2 + 36x \).
\( f(x)=x(2x^{3}+9x^{2}+8x + 36) \)

Step2: Group the terms in the cubic polynomial

Now, we group the terms in the cubic polynomial \( 2x^{3}+9x^{2}+8x + 36 \) as follows:
\( (2x^{3}+9x^{2})+(8x + 36) \)

Step3: Factor out the common factors from each group

Factor out \( x^{2} \) from the first group and \( 4 \) from the second group:
\( x^{2}(2x + 9)+4(2x + 9) \)

Step4: Factor out the common binomial factor

Now, we can factor out the common binomial factor \( (2x + 9) \) from the two terms:
\( (2x + 9)(x^{2}+4) \)

Step5: Combine the factors

Putting it all together, the factored form of the function is:
\( f(x)=x(2x + 9)(x^{2}+4) \)

Step6: Find the x - intercepts

To find the \( x \)-intercepts, we set \( f(x)=0 \). So we have the equation:
\( x(2x + 9)(x^{2}+4)=0 \)
We set each factor equal to zero and solve for \( x \):

• For \( x = 0 \), the equation is satisfied.
• For \( 2x+9 = 0 \), we solve for \( x \): \( 2x=-9\Rightarrow x =-\frac{9}{2} \)
• For \( x^{2}+4=0 \), we have \( x^{2}=- 4 \). Since the square of a real number cannot be negative, there are no real solutions from this factor. But we are looking for \( x \)-intercepts (real - valued), so we consider the real solutions from the other factors.

Answer:

The \( x \)-intercepts of the function are \( x = 0 \) and \( x=-\frac{9}{2} \) (or \( x = 0 \) and \( x=-4.5\)). In boxed form, the \( x \)-intercepts are \( \boxed{0} \) and \( \boxed{-\dfrac{9}{2}} \)