Question

multiply. write your answer in simplest form. \\(\frac{35s - 28}{5s - 4} \cdot \frac{4s^2 - 4s + 1}{40s^3 + 8s^2}\\)

Explanation:

Step1: Factor numerators and denominators

First, factor the numerator of the first fraction: \(35s - 28 = 7(5s - 4)\).
Factor the numerator of the second fraction: \(4s^2 - 4s + 1=(2s - 1)^2\) (using the perfect square formula \(a^2-2ab + b^2=(a - b)^2\), here \(a = 2s\), \(b = 1\)).
Factor the denominator of the second fraction: \(40s^3+8s^2 = 8s^2(5s + 1)\)? Wait, no, let's check again. Wait, \(40s^3+8s^2=8s^2(5s + 1)\)? No, wait, original denominator is \(40s^3 + 8s^2\), factor out \(8s^2\)? Wait, no, wait the first denominator is \(5s - 4\), second denominator: \(40s^3+8s^2 = 8s^2(5s + 1)\)? Wait, no, maybe I made a mistake. Wait, let's re - factor:

Wait, the second fraction's denominator: \(40s^3+8s^2=8s^2(5s + 1)\)? No, wait, \(40s^3+8s^2 = 8s^2(5s + 1)\)? Wait, no, actually, \(40s^3+8s^2=8s^2(5s + 1)\)? Wait, no, let's do it step by step. The second fraction is \(\frac{4s^2 - 4s + 1}{40s^3+8s^2}\). Let's factor numerator and denominator:

Numerator: \(4s^2-4s + 1=(2s - 1)^2\) (since \((2s-1)^2=4s^2-4s + 1\))

Denominator: \(40s^3+8s^2=8s^2(5s + 1)\)? Wait, no, \(40s^3+8s^2 = 8s^2(5s + 1)\)? Wait, \(40s^3\div8s^2 = 5s\), \(8s^2\div8s^2 = 1\), so yes, \(40s^3 + 8s^2=8s^2(5s + 1)\). Wait, but the first fraction's numerator is \(35s - 28=7(5s - 4)\), denominator is \(5s - 4\).

So the original expression is \(\frac{7(5s - 4)}{5s - 4}\cdot\frac{(2s - 1)^2}{8s^2(5s + 1)}\)

Step2: Cancel common factors

We can cancel out the common factor \((5s - 4)\) from the first fraction's numerator and denominator. So after canceling \((5s - 4)\), we have:

\(7\cdot\frac{(2s - 1)^2}{8s^2(5s + 1)}\)

Wait, wait, no, wait the second fraction's denominator: Wait, maybe I factored the second denominator wrong. Wait, the second fraction's denominator is \(40s^3+8s^2\), let's factor again: \(40s^3+8s^2 = 8s^2(5s + 1)\)? Wait, no, \(40s^3+8s^2=8s^2(5s + 1)\)? Wait, \(40s^3=8s^2\times5s\), \(8s^2=8s^2\times1\), so yes. But wait, the first fraction is \(\frac{35s - 28}{5s - 4}=\frac{7(5s - 4)}{5s - 4}\), so when we multiply by \(\frac{4s^2 - 4s + 1}{40s^3+8s^2}\), we have:

\(\frac{7(5s - 4)}{5s - 4}\times\frac{(2s - 1)^2}{8s^2(5s + 1)}\)

Cancel \((5s - 4)\) from numerator and denominator of the first fraction:

We get \(7\times\frac{(2s - 1)^2}{8s^2(5s + 1)}=\frac{7(2s - 1)^2}{8s^2(5s + 1)}\)

Wait, but let's check the second fraction's numerator again. Wait, \(4s^2-4s + 1=(2s - 1)^2\), that's correct. The first fraction: \(35s - 28 = 7(5s - 4)\), denominator \(5s - 4\), so cancel \(5s - 4\), we get 7. Then multiply by \(\frac{(2s - 1)^2}{40s^3+8s^2}\). Wait, \(40s^3+8s^2=8s^2(5s + 1)\), so:

\(7\times\frac{(2s - 1)^2}{8s^2(5s + 1)}=\frac{7(2s - 1)^2}{8s^2(5s + 1)}\)

Wait, but maybe I made a mistake in factoring the second denominator. Wait, let's re - express the original problem. The original problem is \(\frac{35s - 28}{5s - 4}\cdot\frac{4s^2 - 4s + 1}{40s^3+8s^2}\)

Factor \(35s - 28 = 7(5s - 4)\)

Factor \(4s^2 - 4s + 1=(2s - 1)^2\)

Factor \(40s^3+8s^2=8s^2(5s + 1)\)

So now the expression is \(\frac{7(5s - 4)}{5s - 4}\cdot\frac{(2s - 1)^2}{8s^2(5s + 1)}\)

Cancel \((5s - 4)\) from the first fraction:

\(7\cdot\frac{(2s - 1)^2}{8s^2(5s + 1)}=\frac{7(2s - 1)^2}{8s^2(5s + 1)}\)

Wait, but let's check if there are other common factors. There are no other common factors between the numerator and the denominator. So the simplified form is \(\frac{7(2s - 1)^2}{8s^2(5s + 1)}\) or we can expand \((2s - 1)^2=4s^2-4s + 1\), so \(\frac{7(4s^2-4s + 1)}{8s^2(5s + 1)}=\frac{28s^2-28s + 7}{40s^3+8s^2}\) (but the factored form is simpler)

Wait, maybe I made a mistake in the second denominator. Wait, \(40s^3+8s^2 = 8s^2(5s + 1)\)? Wait, \(40s^3+8s^2=8s^2(5s + 1)\), yes. So the cancellation is correct.

Answer:

\(\boldsymbol{\frac{7(2s - 1)^2}{8s^2(5s + 1)}}\) (or expanded form \(\frac{28s^2-28s + 7}{40s^3+8s^2}\))