Petra is practicing factoring by grouping. She wants to factor the polynomial 16xy−4x^2+28y−7x. Her work is shown below.
16xy−4x^2+28y−7x
4x(4y−x)+7(4y−x)
(4x+7)(4y−x)
Which statements regarding Petra’s work are true?
There may be more than one correct answer. Select all correct answers.
1. The constant 7 is not the greatest common factor of 28y−7x.
2. Petra is correct.
3. The product of Petra’s factors is not the original polynomial.
4. The monomial 4x is not the greatest common factor of 16xy−4x^2.
5. Petra is incorrect.
6. The product of Petra’s factors is the original polynomial.
could somebody help
She is correct!
yes, but you need to choose 2,6 as instructed
To determine which statements regarding Petra's work are true, let's analyze each statement one by one:
1. The constant 7 is not the greatest common factor of 28y−7x.
- This statement is true. The greatest common factor of 28y and -7x is not 7, but rather 1, as both terms can be divided by 7.
2. Petra is correct.
- This statement is true. Petra correctly factored the polynomial using the method of factoring by grouping.
3. The product of Petra’s factors is not the original polynomial.
- This statement is false. Petra's factored form, (4x+7)(4y−x), is indeed the original polynomial when multiplied back out.
4. The monomial 4x is not the greatest common factor of 16xy−4x^2.
- This statement is true. The greatest common factor of 16xy and -4x^2 is not 4x, but rather -4x, as both terms can be divided by -4x.
5. Petra is incorrect.
- This statement is false. Petra's work is correct.
6. The product of Petra’s factors is the original polynomial.
- This statement is true. The product of Petra's factors, (4x+7)(4y−x), does indeed equal the original polynomial, 16xy−4x^2+28y−7x.
Therefore, the correct statements are:
- Statement 1: The constant 7 is not the greatest common factor of 28y−7x.
- Statement 2: Petra is correct.
- Statement 6: The product of Petra’s factors is the original polynomial.