Questions: Factor the trinomial completely.
28 x^3 + 51 x^2 + 11 x
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. 28 x^3 + 51 x^2 + 11 x = (Factor completely.)
B. The polynomial is prime.
Transcript text: 02 Intermediate Algebra (2253-1)
k: IA 13.3
on
Objective A
Objective B
Objective C
tive B: Factor Out a GCF Before Factoring a Trinomial of the Form $a x^{2}+b x+c$
UR TURN \#1
0 of 1 Point
Factor the trinomial completely.
\[
28 x^{3}+51 x^{2}+11 x
\]
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $28 x^{3}+51 x^{2}+11 x=\square$ (Factor completely.)
$\square$
B. The polynomial is prime.
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Solution
Solution Steps
Step 1: Identify the Greatest Common Factor (GCF)
First, identify the GCF of the terms \( 28x^{3} \), \( 51x^{2} \), and \( 11x \). The GCF of the coefficients \( 28 \), \( 51 \), and \( 11 \) is \( 1 \). The GCF of the variable terms \( x^{3} \), \( x^{2} \), and \( x \) is \( x \). Therefore, the GCF of the entire expression is \( x \).
Step 2: Factor Out the GCF
Factor out the GCF \( x \) from the trinomial:
\[
28x^{3} + 51x^{2} + 11x = x(28x^{2} + 51x + 11)
\]
Step 3: Factor the Remaining Quadratic Expression
Now, factor the quadratic expression \( 28x^{2} + 51x + 11 \). To do this, find two numbers that multiply to \( 28 \times 11 = 308 \) and add to \( 51 \). The numbers \( 44 \) and \( 7 \) satisfy this condition because \( 44 \times 7 = 308 \) and \( 44 + 7 = 51 \).
Rewrite the middle term using these numbers:
\[
28x^{2} + 44x + 7x + 11
\]
Group the terms:
\[
(28x^{2} + 44x) + (7x + 11)
\]
Factor out the GCF from each group:
\[
4x(7x + 11) + 1(7x + 11)
\]
Factor out the common binomial factor \( (7x + 11) \):
\[
(7x + 11)(4x + 1)
\]
Step 4: Write the Final Factored Form
Combine the GCF \( x \) with the factored quadratic expression:
\[
28x^{3} + 51x^{2} + 11x = x(7x + 11)(4x + 1)
\]