Respuesta :
Answer:
Both are correct.
Step-by-step explanation:
The key understanding here is that you can factor a monomial in many different ways!
To check if any of the factorizations is correct, we can multiply the factors and see if their product is really 12x^712x
7
12, x, start superscript, 7, end superscript.
Hint #22 / 4
\begin{aligned} (\blueD{4}\maroonD{x^3})(\blueD{3}\maroonD{x^4})&=(\blueD{4})(\blueD{3})(\maroonD{x^3})(\maroonD{x^4}) \\\\ &=\blueD{12}\maroonD{x^7} \end{aligned}
(4x
3
)(3x
4
)
=(4)(3)(x
3
)(x
4
)
=12x
7
So Ibuki is correct!
Hint #33 / 4
\begin{aligned} (\blueD{2}\maroonD{x^6})(\blueD{6}\maroonD{x})&=(\blueD{2})(\blueD{6})(\maroonD{x^6})(\maroonD{x}) \\\\ &=\blueD{12}\maroonD{x^7} \end{aligned}
(2x
6
)(6x)
=(2)(6)(x
6
)(x)
=12x
7
So Melodie is also correct!
Both Ibuki and Melodie are correct.
Both Ibuki and Melodie have correctly factored [tex]12x^{7}[/tex].
What is meant by factor?
A number or algebraic expression that divides another number or expression evenly is called factor.
How to find Ibuki and Melodie factored 12x^7 correctly?
According to the problem,
- Ibuki factored [tex]12x^{7}[/tex] as (4x^3)(3x^4).
This is absolutely correct because the terms (4x^3) and (3x^4) if multiplied gives [tex]12x^{7}[/tex].
And if any of these terms divides the term [tex]12x^{7}[/tex] gives the other terms as quotient with no remainder.
- Melodie factored [tex]12x^{7}[/tex] as (2x^6) (6x).
This is absolutely correct because the terms (2x^6) and 6x if multiplied gives [tex]12x^{7}[/tex].
And if any of these terms divides the term [tex]12x^{7}[/tex] gives the other terms as quotient with no remainder.
From here we can conclude that Both Ibuki and Melodie have correctly factored [tex]12x^{7}[/tex].
Find more about "Factor" here: https://brainly.com/question/219464
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