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david found and factored out the gcf of the polynomial 80b⁴ - 32b²c³ + …

Question

david found and factored out the gcf of the polynomial 80b⁴ - 32b²c³ + 48b⁴c. his work is below. 1. gcf of 80, 32, and 48: 16 2. gcf of b⁴, b², and b⁴: b² 3. gcf of c³ and c: c 4. gcf of the polynomial: 16b²c 5. rewrite as a product of the gcf: 16b²c(5b²) - 16b²c(2c²) + 16b²c(3b²) 6. factor out gcf: 16b²c(5b² - 2c² + 3b²) which statements are true about david’s work? check all that apply. ☐ the gcf of the coefficients is correct. ☐ the gcf of the variable b should be b⁴ instead of b². ☐ the variable c is not common to all terms, so a power of c should not have been factored out. ☐ the expression in step 5 is equivalent to the given polynomial. ☐ in step 6, david applied the distributive property.

Explanation:

Brief Explanations
  1. For coefficients 80, 32, 48: The greatest common factor is 16, so the first statement is true.
  2. For variable $b$: The terms have $b^4$, $b^2$, $b^4$; the lowest power is $b^2$, so the second statement is false.
  3. For variable $c$: The first term $80b^4$ has no $c$, so $c$ is not common to all terms and should not be in the GCF, making the third statement true.
  4. Step 5: The first term $16b^2c(5b^2)=80b^4c$, but the original first term is $80b^4$, so they are not equivalent, making the fourth statement false.
  5. Step 6: Factoring out the GCF uses the reverse distributive property, so the fifth statement is true.

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.