Question

vp +12v +8p -96

Explanation:

Step1: Group like terms

We have the expression \( vp + 12v+8p - 96\). Group terms with \(v\) and terms with \(p\) (and constants): \((vp + 12v)+(8p - 96)\).

Step2: Factor out common factors

From the first group \(vp + 12v\), factor out \(v\): \(v(p + 12)\). From the second group \(8p - 96\), factor out 8: \(8(p - 12)\)? Wait, no, \(8p-96 = 8(p - 12)\)? Wait, no, \(96\div8 = 12\), so \(8p-96=8(p - 12)\)? Wait, but the first group is \(v(p + 12)\). Wait, maybe I made a mistake. Wait, let's re - group. Maybe group \(vp+8p\) and \(12v - 96\).

Step3: Re - group and factor

Group as \((vp + 8p)+(12v - 96)\). Factor \(p\) from the first group: \(p(v + 8)\). Factor 12 from the second group: \(12(v - 8)\). No, that's not helpful. Wait, maybe the original expression is \(vp+12v + 8p+96\)? Wait, the user wrote \(-96\). Wait, maybe it's a factoring by grouping problem. Let's try again.

Wait, let's assume the expression is \(vp + 12v+8p + 96\) (maybe a sign error). Then:

Group \((vp + 12v)+(8p + 96)=v(p + 12)+8(p + 12)=(v + 8)(p + 12)\). But with \(-96\), let's see:

\(vp+12v + 8p-96=v(p + 12)+8(p - 12)\). No, that doesn't factor nicely. Wait, maybe the original expression is \(vp + 12v+8p + 96\). Let's proceed with the factoring by grouping.

If we have \(vp+12v + 8p + 96\):

1. Group terms: \((vp + 12v)+(8p + 96)\)
2. Factor: \(v(p + 12)+8(p + 12)\)
3. Factor out \((p + 12)\): \((v + 8)(p + 12)\)

But if the expression is \(vp+12v + 8p-96\), let's check the arithmetic. \(12v-96 = 12(v - 8)\), \(vp + 8p=p(v + 8)\). So it's \(p(v + 8)+12(v - 8)\), which doesn't factor into a product of two binomials with integer coefficients. Maybe there was a typo, but assuming the expression is \(vp + 12v+8p + 96\), the factored form is \((v + 8)(p + 12)\).

Answer:

If the expression is \(vp + 12v+8p + 96\), the factored form is \(\boldsymbol{(v + 8)(p + 12)}\). If the expression is \(vp + 12v+8p-96\), it can be written as \(v(p + 12)+8(p - 12)\) (or with further manipulation, but it's less clean).