Question

teresa is factoring this polynomial by grouping. which common factors should be used in the next step of factoring?\\(10x^{3}+3x^{2}-20x-6\\)\\((10x^{3}+3x^{2})+(-20x-6)\\)\\(\bigcirc\\ x^{2}\\) and \\(-2x\\)\\(\bigcirc\\ 2x^{2}\\) and \\(-2x\\)\\(\bigcirc\\ x^{2}\\) and \\(-2\\)\\(\bigcirc\\ 2x^{2}\\) and \\(-2\\)

Explanation:

Step1: Factor first group

For \(10x^{3}+3x^{2}\), find GCF. The GCF of \(10x^{3}\) and \(3x^{2}\) is \(x^{2}\)? Wait, no, wait: \(10x^{3}=x^{2}\times10x\), \(3x^{2}=x^{2}\times3\)? Wait, no, wait, maybe I miscalculated. Wait, \(10x^{3}+3x^{2}\): factors of 10 and 3, GCF of 10 and 3 is 1? No, wait, the first group is \(10x^{3}+3x^{2}\), the GCF of coefficients 10 and 3 is 1? No, wait, maybe I made a mistake. Wait, the second group is \(-20x - 6\), GCF of -20 and -6 is -2. Wait, no, let's re - examine.

Wait, the first group: \(10x^{3}+3x^{2}\). Let's factor out \(x^{2}\): \(x^{2}(10x + 3)\). The second group: \(-20x-6\). Let's factor out -2: \(-2(10x + 3)\). Wait, so for the first group, the common factor is \(x^{2}\), for the second group, the common factor is -2.

Let's check the options:

Option 1: \(x^{2}\) and \(-2x\): If we factor out \(-2x\) from \(-20x - 6\), we get \(-2x(10+\frac{3}{x})\), which is not a polynomial factorization, so this is wrong.

Option 2: \(2x^{2}\) and \(-2x\): Factoring out \(2x^{2}\) from \(10x^{3}+3x^{2}\) gives \(2x^{2}(5x+\frac{3}{2})\), not a polynomial, so wrong.

Option 3: \(x^{2}\) and \(-2\): Factor out \(x^{2}\) from \(10x^{3}+3x^{2}\): \(x^{2}(10x + 3)\). Factor out -2 from \(-20x-6\): \(-2(10x + 3)\). This works.

Option 4: \(2x^{2}\) and \(-2\): Factoring out \(2x^{2}\) from \(10x^{3}+3x^{2}\) gives \(2x^{2}(5x+\frac{3}{2})\), not a polynomial, so wrong.

Step 2: Conclusion

So the common factors are \(x^{2}\) and \(-2\).

Answer:

\(x^{2}\) and \(-2\) (corresponding to the option " \(x^{2}\) and \(-2\)")