Question

1. f(x) = 2x³ - 3x² - 32x + 48

factored form:
solutions:

Explanation:

Step1: Group the terms

We have the polynomial \( f(x) = 2x^{3}-3x^{2}-32x + 48 \). Group the first two terms and the last two terms:
\( f(x)=(2x^{3}-3x^{2})+(-32x + 48) \)

Step2: Factor out the GCF from each group

From the first group \( 2x^{3}-3x^{2} \), the greatest common factor (GCF) is \( x^{2} \), so we factor that out: \( x^{2}(2x - 3) \)
From the second group \( -32x + 48 \), the GCF is \( -16 \), so we factor that out: \( -16(2x - 3) \)
Now the polynomial becomes: \( f(x)=x^{2}(2x - 3)-16(2x - 3) \)

Step3: Factor out the common binomial factor

Notice that both terms have a common factor of \( (2x - 3) \). Factor that out:
\( f(x)=(2x - 3)(x^{2}-16) \)

Step4: Factor the difference of squares

The term \( x^{2}-16 \) is a difference of squares, since \( x^{2}-16=x^{2}-4^{2} \). Using the difference of squares formula \( a^{2}-b^{2}=(a + b)(a - b) \), where \( a = x \) and \( b = 4 \), we get:
\( x^{2}-16=(x + 4)(x - 4) \)
So the factored form of the polynomial is \( f(x)=(2x - 3)(x + 4)(x - 4) \)

Step5: Find the solutions

To find the solutions, we set \( f(x)=0 \):
\( (2x - 3)(x + 4)(x - 4)=0 \)
Using the zero - product property, which states that if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \) (or both), we set each factor equal to zero:

• For \( 2x-3 = 0 \), we solve for \( x \): \( 2x=3\Rightarrow x=\frac{3}{2} \)
• For \( x + 4=0 \), we solve for \( x \): \( x=-4 \)
• For \( x - 4=0 \), we solve for \( x \): \( x = 4 \)

Answer:

Factored Form: \( (2x - 3)(x + 4)(x - 4) \)
Solutions: \( x=\frac{3}{2},x=-4,x = 4 \)