Algebra Part 2 Activity

Section 2.2 Algebra Part 2 Activity

Subsection 2.2.1 Factoring Out a Variable

Activity 2.2.1.

Solve by factoring out a variables and cancelling. Maybe sure to note any extra solutions produced by cancelling.

\begin{equation*} 9x^2 + 3x = 12x^2 \end{equation*}

Solution

We factor the variable \(x\) out of each side.

\begin{equation*} x(9x + 3) = x(12x) \end{equation*}

Then we can cancel of (or equivalently divide by) the variable.

\begin{equation*} 9x + 3 = 12x \end{equation*}

Then we solve. Here, I'll subtract \(9x\) from both sides of the equation.

\begin{equation*} 3 = 3x \end{equation*}

Dividing by 3 gives the solution \(x=1\text{.}\) Since we divided by the variable, we must also consider the possibility that \(x=0\text{.}\) If we put \(x=0\) in the origina equation, we get \(0=0\text{,}\) which is true, so \(x=0\) is a second solution.

Activity 2.2.2.

Solve by factoring out a variables and cancelling. Maybe sure to note any extra solutions produced by cancelling.

\begin{equation*} 7t^2 + 8t^3 = 3t^2 \end{equation*}

Solution

Here the lowest power of the variable is \(t^2\text{,}\) so we can factor that out of both sides of the equation.

\begin{equation*} t^2(7 + 8t) = t^2(3) \end{equation*}

Then we can cancel off (or equivalently divide by) \(t^2\text{.}\)

\begin{equation*} 7 + 8t = 3 \end{equation*}

Then we solve by subtracting 7 from both sides of the equation.

\begin{equation*} 8t = -4 \end{equation*}

Finally, we divide by 8.

\begin{equation*} t = \frac{-4}{8} = \frac{-1}{2} \end{equation*}

Again, since we divided by \(t^2\text{,}\) we must consider the possibility that \(t=0\text{.}\) This value satisafy the original equation, so we conclude that \(t=0\) is a second solution.

Activity 2.2.3.

Solve by factoring out a variables and cancelling. Maybe sure to note any extra solutions produced by cancelling.

\begin{equation*} 4l^3 + 8l = 20l \end{equation*}

Solution

We can factor out \(l\) from both sides of the equation.

\begin{equation*} l(4l^2 + 8) = l(20) \end{equation*}

We can cancel off (or equivalently divide by) \(l\text{.}\)

\begin{equation*} 4l^2 + 8 = 20 \end{equation*}

Then we can subtract 8 from both sides of the equation.

\begin{equation*} 4l^2 = 12 \end{equation*}

Then we can divide by 4.

\begin{equation*} l^2 = 3 \end{equation*}

Finally, to isolate the variable, we have to take the square root of both sides.

\begin{equation*} l = \pm \sqrt{3} \end{equation*}

The notation \(\sqrt{3}\) is implicity positive, but we could also have the variable be the negative square root. Therefore, we write \(\pm \sqrt{3}\) for both possible solutions.

Subsection 2.2.2 Factoring with Square Roots

Activity 2.2.4.

Simplify this square root by factoring out a factor which a reasonable square root.

\begin{equation*} \sqrt{150} \end{equation*}

Solution

We need to find a factor of 302 which has a nice square root. We can do this by trial an error, trying small square numbers (trying factors of 4, 9, 16, 25, and so on). We would first find that 25 is s a factor which has a nice square root.

\begin{equation*} \sqrt{150} = \sqrt{(25)(6)} \end{equation*}

Then we can split up the square root.

\begin{equation*} \sqrt{25} \sqrt{6} \end{equation*}

Then we can evaluate the square root of 25.

\begin{equation*} 5\sqrt{6} \end{equation*}

Activity 2.2.5.

Simplify this square root by factoring out a factor which a reasonable square root.

\begin{equation*} \sqrt{343} \end{equation*}

Solution

We need to find a factor of 392 which has a nice square root. We can do this by trial and error, trying small values. We find that 49 is a factor which has a nice square root. (Again, I just did this by testing small factors with nice square roots: 4,, 9, 16, and so on).

\begin{equation*} \sqrt{343} = \sqrt{(49)(7)} \end{equation*}

Then we split up the square root.

\begin{equation*} \sqrt{49}\sqrt{7} \end{equation*}

Then we evalute the square root of 49.

\begin{equation*} 7 \sqrt{7} \end{equation*}

Activity 2.2.6.

Factor a variable out of this square root expression.

\begin{equation*} \sqrt{9z^2 + 13z^3} \end{equation*}

Solution

We can see that \(z^2\) is a factor of both terms in the square root, so we factor that out.

\begin{equation*} \sqrt{z^2(9 + 13z)} \end{equation*}

Then we can split up the square root.

\begin{equation*} \sqrt{z^2}\sqrt{9 + 13z} \end{equation*}

Then we can evalute the first square root.

\begin{equation*} |z| \sqrt{9 + 13z} \end{equation*}

Now we've factor out a variable and simplified the square root.

Activity 2.2.7.

Factor a variable out of this square root expression.

\begin{equation*} \sqrt{u^4 - u^3} \end{equation*}

Solution

The term \(u^3\) is a factor of both terms under the square root. However \(u^3\) isn't a term we can reasonable take out of a square root: there isn't a nice square root of a cube. Instead, I'll factor \(u^2\) out of both terms.

\begin{equation*} \sqrt{u^2 (u^2 - u)} \end{equation*}

Then we can split up the square root.

\begin{equation*} \sqrt{u^2} \sqrt{u^2 - u} \end{equation*}

Then we can evalute the first square root.

\begin{equation*} |u| \sqrt{u^2 - u} \end{equation*}

Now we've factored a variable out of the square root and simplified the square root expression.

Subsection 2.2.3 Conceputal Questions

Activity 2.2.8.

What does it mean to factor?
What is the distribute law and where does it apply?