Question

david found and factored out the gcf of the polynomial $80b^4 - 32b^2c^3 + 48b^4c$. his work is below.

1. gcf of 80, 32, and 48: 16
2. gcf of $b^4$, $b^2$, and $b^4$: $b^2$
3. gcf of $c^3$ and $c$: $c$
4. gcf of the polynomial: $16b^2c$
5. rewrite as a product of the gcf:

$16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2)$

1. factor out gcf: $16b^2c(5b^2 - 2c^2 + 3b^2)$

which statements are true about davids work? check all that apply.
$square$ the gcf of the coefficients is correct.
$square$ the gcf of the variable $b$ should be $b^4$ instead of $b^2$.
$square$ the variable $c$ is not common to all terms, so a power of $c$ should not have been factored out.
$square$ the expression in step 5 is equivalent to the given polynomial.
$square$ in step 6, david applied the distributive property.

Explanation:

Step1: Verify coefficient GCF

GCF of 80,32,48: $\gcd(80,32,48)=16$, so this is correct.

Step2: Verify variable $b$ GCF

Terms have $b^4, b^2, b^4$. The lowest power is $b^2$, so David's work is correct here.

Step3: Verify variable $c$ GCF

First term $80b^4$ has no $c$, so $c$ is not common to all terms. Factoring out $c$ is wrong.

Step4: Check step5 equivalence

Calculate $16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2) = 80b^4c - 32b^2c^3 + 48b^4c$, which is not equal to original $80b^4 - 32b^2c^3 + 48b^4c$.

Step5: Check step6 property

Factoring out the GCF uses the distributive property $ab+ac=a(b+c)$, so this is correct.

Answer:

• The GCF of the coefficients is correct.
• The variable $c$ is not common to all terms, so a power of $c$ should not have been factored out.
• In step 6, David applied the distributive property.