Question

1. $3x^2 - 27$ $(3x+$

Explanation:

Step1: Factor out the GCF

First, factor out the greatest common factor (GCF) of \(3x^2\) and \(-27\), which is \(3\). So we have:
\(3x^2 - 27 = 3(x^2 - 9)\)

Step2: Apply difference of squares

Notice that \(x^2 - 9\) is a difference of squares, since \(x^2 - 9 = x^2 - 3^2\). The formula for the difference of squares is \(a^2 - b^2=(a + b)(a - b)\). Here, \(a = x\) and \(b = 3\), so:
\(x^2 - 9=(x + 3)(x - 3)\)

Step3: Combine with the GCF

Substituting back into the expression from Step1, we get:
\(3x^2 - 27 = 3(x + 3)(x - 3)\)
If we want to factor it in the form with \(3x\) as a term, we can also factor it as follows (maybe there was a typo in the given form, but let's adjust):
Wait, the given form is \((3x+\dots)\), but actually the correct factoring with GCF and difference of squares is \(3(x + 3)(x - 3)\). If we expand \(3(x + 3)(x - 3)\), we get \(3(x^2 - 9)=3x^2 - 27\). But if we try to factor it as \((3x + a)(x + b)\), let's multiply \((3x + a)(x + b)=3x^2+(3b + a)x+ab\). We want this to equal \(3x^2 - 27\), so the \(x\) term coefficient must be \(0\) and \(ab=-27\). Let's solve:
\(3b + a = 0\) and \(ab=-27\)
From \(3b + a = 0\), we have \(a=-3b\). Substitute into \(ab=-27\):
\(-3b\times b=-27\)
\(-3b^2=-27\)
\(b^2 = 9\)
\(b = 3\) or \(b=-3\)
If \(b = 3\), then \(a=-9\). Let's check \((3x - 9)(x + 3)=3x^2+9x - 9x - 27=3x^2 - 27\). Yes, that works. So \((3x - 9)(x + 3)=3(x - 3)(x + 3)\), which is the same as before.
But the given form is \((3x+\dots)\), maybe there was a mistake in the problem's expected form. However, the correct factoring of \(3x^2 - 27\) is \(3(x + 3)(x - 3)\) or \((3x - 9)(x + 3)\) (after factoring out 3 from the first binomial: \(3(x - 3)(x + 3)=(3x - 9)(x + 3)\)).

Answer:

The factored form of \(3x^2 - 27\) is \(3(x + 3)(x - 3)\) (or \((3x - 9)(x + 3)\) if following the given binomial start with \(3x\)). If we consider the given partial form \((3x+\dots)\), the complete factoring with that start would be \((3x - 9)(x + 3)\) (since \((3x - 9)(x + 3)=3x^2+9x - 9x - 27=3x^2 - 27\)). So the missing term in \((3x+\dots)\) is \(-9\) and the other binomial is \((x + 3)\), so the full factoring is \((3x - 9)(x + 3)\) or simplified as \(3(x + 3)(x - 3)\).