Q. Express as a trinomial: \( (2x – 8)(2x – 6) = 4x^2 – 28x + 48 \).

Q: How do I expand (2x-8)(2x-6) using FOIL?

FOIL: First 4x^2, Outer -12x, Inner -16x, Last +48. Combine like terms to get 4x^2-28x+48.

Q: Can I factor common factors before multiplying?

Yes. Factor 2 from each: 2(x-4)\cdot2(x-3)=4(x-4)(x-3). Expanding yields 4x^2-28x+48..

Q: Is the product a trinomial?

Yes. The expanded form 4x^2-28x+48 is a quadratic trinomial (three terms, degree 2).

Q: What are the roots/zeros of the trinomial?

Divide by 4: x^2-7x+12=(x-3)(x-4). Roots are x=3 and x=4.

Q: Can the trinomial be factored further?

Yes: 4x^2-28x+48=4(x-3)(x-4), which is the fully factored form.

Q: What is the axis of symmetry and vertex?

Axis: x=-\tfrac{b}{2a}=\tfrac{28}{8}=3.5. Vertex: evaluate gives (3.5,-1)..

Q: What is the y-intercept?

Set x=0: y=48. The y-intercept is 48.

Answer

\[
\begin{aligned}
(2x-8)(2x-6) &= 2x \cdot 2x + 2x \cdot (-6) + (-8) \cdot 2x + (-8) \cdot (-6) \\
&= 4x^2 – 12x – 16x + 48 = 4x^2 – 28x + 48
\end{aligned}
\]

Detailed Explanation

Write the original expression to be expanded.\[ (2x – 8)(2x – 6) \]
Factor out the greatest common factor 2 from each binomial to simplify multiplication. Do this separately for each parenthesis:\(2x – 8 = 2(x – 4)\)\(2x – 6 = 2(x – 3)\)
Replace the original expression with the factored forms and multiply the numeric factors together. Explain the operation separately:Numeric multiplication: \(2 \times 2 = 4\)Therefore the product becomes:
\[ (2x – 8)(2x – 6) = 4\,(x – 4)(x – 3) \]
Now expand the quadratic factor \((x – 4)(x – 3)\) using FOIL (multiply First, Outer, Inner, Last) and state each partial product separately:First: \(x \times x = x^{2}\)Outer: \(x \times (-3) = -3x\)
Inner: \((-4) \times x = -4x\)

Last: \((-4) \times (-3) = 12\)

Add these results to get the quadratic trinomial:

\[ (x – 4)(x – 3) = x^{2} – 3x – 4x + 12 = x^{2} – 7x + 12 \]
Multiply this trinomial by the numeric factor 4. Do the multiplication for each term separately:Multiply \(4 \times x^{2} = 4x^{2}\)Multiply \(4 \times (-7x) = -28x\)
Multiply \(4 \times 12 = 48\)

Combine these results to obtain the final trinomial:

\[ 4(x^{2} – 7x + 12) = 4x^{2} – 28x + 48 \]
State the final answer (the expression written as a trinomial):\[ \boxed{4x^{2} – 28x + 48} \]

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Algebra FAQs

How do I expand \( (2x-8)(2x-6) \) using FOIL?

FOIL: First \(4x^2\), Outer \(-12x\), Inner \(-16x\), Last \(+48\). Combine like terms to get \(4x^2-28x+48\).

Can I factor common factors before multiplying?

Yes. Factor 2 from each: \(2(x-4)\cdot2(x-3)=4(x-4)(x-3)\). Expanding yields \(4x^2-28x+48\)..

Is the product a trinomial?

Yes. The expanded form \(4x^2-28x+48\) is a quadratic trinomial (three terms, degree 2).

What are the roots/zeros of the trinomial?

Divide by 4: \(x^2-7x+12=(x-3)(x-4)\). Roots are \(x=3\) and \(x=4\).

Can the trinomial be factored further?

Yes: \(4x^2-28x+48=4(x-3)(x-4)\), which is the fully factored form.

What is the leading coefficient and degree?

What is the axis of symmetry and vertex?

Axis: \(x=-\tfrac{b}{2a}=\tfrac{28}{8}=3.5\). Vertex: evaluate gives \((3.5,-1)\)..

What is the \(y\)-intercept?

Set \(x=0\): \(y=48\). The \(y\)-intercept is \(48\).

Turn (2x-8)(2x-6) into the trinomial.
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