David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below.
2 answers:
Answer:
Given Polynomial:
Factors of Coefficient of terms
80 = 5 × 16
32 = 2 × 16
48 = 3 × 16
Common factor of the coefficient of all term is 16.
Each term contain variable. So the Minimum power of b is common from all terms.
Common from all variable part comes b².
So, Common factor of the polynomial = 16b²
⇒ 16b² ( 5b² ) - 16b² ( 2c³ ) + 16b² ( 3b²c )
⇒ 16b² ( 5b² - 2c³ + 3b²c )
Therefore, Statements that are true about David's word are:
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
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.
The expression in step 5 is equivalent to the given polynomial.
In step 6, David applied the distributive property
Step-by-step explanation:
Only the above four statements are true.
GFC of 80, 32, and 48: 16
We find that 16 is the highest number which divides 80, 32 and 48. Hence 16 is GFC.
GCF of b4, b2, and b4: b2
Also we have taken the term of b in all three and found the least exponent as gCF hence correct.
Since c is not in the first term, no c term can be GCF hence iii is also true.
But the expression in step 5 is not equivalent to the given polynomial because I term in the step 5 = , but in the given no c term is there.
Yes. In step 6, he applied distributive property to take GCF outside the polynomial as a factor.
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