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Which of the following worked equations demonstrates that you can divide polynomials by recognizing division as the inverse ope...Question
Which of the following worked equations demonstrates that you can divide polynomials by recognizing division as the inverse operation of multiplication?(1 point) Responses 8x2−4x+124x=(4x)(8x2−4x+12) Start Fraction 8 x squared minus 4 x plus 12 over 4 x End Fraction equals left parenthesis 4 x right parenthesis left parenthesis 8 x squared minus 4 x plus 12 right parenthesis 8x2−4x+124x=(−14x)(8x2−4x+12) Start Fraction 8 x squared minus 4 x plus 12 over 4 x End Fraction equals left parenthesis negative Start Fraction 1 over 4 x End Fraction right parenthesis left parenthesis 8 x squared minus 4 x plus 12 right parenthesis 8x2−4x+124x=(14x)(8x2−4x+12) Start Fraction 8 x squared minus 4 x plus 12 over 4 x End Fraction equals left parenthesis Start Fraction 1 over 4 x End Fraction right parenthesis left parenthesis 8 x squared minus 4 x plus 12 right parenthesis 8x2−4x+124x=(−4x)(8x2−4x+12)
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Answer
What does the fraction look like
Answered by
GPT-4o mini
To demonstrate that you can divide polynomials by recognizing division as the inverse operation of multiplication, you need an equation that shows that dividing a polynomial by a term is equivalent to finding a factor of that polynomial.
Here’s the polynomial division presented:
Given:
\[ \frac{8x^2 - 4x + 12}{4x} \]
You're looking for an equation that shows that:
\[ \frac{8x^2 - 4x + 12}{4x} = k(8x^2 - 4x + 12) \]
where \( k \) is some constant.
From the provided responses, the equivalent form of the equation should be structured correctly to show this division. The best option is:
1. \[ \frac{8x^2 - 4x + 12}{4x} = \left(\frac{1}{4x}\right)(8x^2 - 4x + 12) \]
This choice correctly represents the division by expressing the polynomial as a multiplication of a constant and a polynomial, confirming the inverse relationship between division and multiplication.
So, the correct response is:
\[ \frac{8x^2 - 4x + 12}{4x} = \left(\frac{1}{4x}\right)(8x^2 - 4x + 12) \]
Here’s the polynomial division presented:
Given:
\[ \frac{8x^2 - 4x + 12}{4x} \]
You're looking for an equation that shows that:
\[ \frac{8x^2 - 4x + 12}{4x} = k(8x^2 - 4x + 12) \]
where \( k \) is some constant.
From the provided responses, the equivalent form of the equation should be structured correctly to show this division. The best option is:
1. \[ \frac{8x^2 - 4x + 12}{4x} = \left(\frac{1}{4x}\right)(8x^2 - 4x + 12) \]
This choice correctly represents the division by expressing the polynomial as a multiplication of a constant and a polynomial, confirming the inverse relationship between division and multiplication.
So, the correct response is:
\[ \frac{8x^2 - 4x + 12}{4x} = \left(\frac{1}{4x}\right)(8x^2 - 4x + 12) \]
Answered by
GPT-4o mini
When representing the expression \(\frac{8x^2 - 4x + 12}{4x}\) as a fraction, it looks like this:
\[
\frac{8x^2 - 4x + 12}{4x}
\]
In this representation:
- The numerator is \(8x^2 - 4x + 12\).
- The denominator is \(4x\).
If you're summing it up in the context of showing the inverse relationship between multiplication and division, this fraction signifies that dividing the polynomial \(8x^2 - 4x + 12\) by \(4x\) can be expressed as multiplying \( (8x^2 - 4x + 12) \) by \(\frac{1}{4x}\):
\[
\frac{8x^2 - 4x + 12}{4x} = \left(\frac{1}{4x}\right)(8x^2 - 4x + 12)
\]
This shows that dividing by \(4x\) is the same as multiplying by \(\frac{1}{4x}\). The fraction is laid out clearly with the polynomial over the term.
\[
\frac{8x^2 - 4x + 12}{4x}
\]
In this representation:
- The numerator is \(8x^2 - 4x + 12\).
- The denominator is \(4x\).
If you're summing it up in the context of showing the inverse relationship between multiplication and division, this fraction signifies that dividing the polynomial \(8x^2 - 4x + 12\) by \(4x\) can be expressed as multiplying \( (8x^2 - 4x + 12) \) by \(\frac{1}{4x}\):
\[
\frac{8x^2 - 4x + 12}{4x} = \left(\frac{1}{4x}\right)(8x^2 - 4x + 12)
\]
This shows that dividing by \(4x\) is the same as multiplying by \(\frac{1}{4x}\). The fraction is laid out clearly with the polynomial over the term.
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