Algebra 2 Standard 4.2 - Factor and solve polynomials Date $$ 2 / 24 / 25 $$ 1) Factor the polynomial completely: $$ x^{3}+x^{2}+4 x+4 $$ Daily Pract 2) Solve: $$ y^{3}-7 y^{2}-4 y+28=0 $$ 3) Solve: $$ \quad n^{3}+5 n^{2}-9 n=45 $$ 4) Factor the polynomial completely: $$ x^{4}-25 $$ $$ \begin{array}{l} \text { let } x^{4}=a^{2}\left\{\begin{array}{l} x^{4}-a 5 \ x^{2}-25 \ x^{2}-25 \ (a-5)(a+5) \ \left(x^{2}-5\right)\left(x^{2}+5\right) \end{array}\right. \end{array} $$ 5) Factor the polynomial completely: $$ a^{4}+5 a^{2}-6 $$ 6) Solve: $$ \quad 3 s^{4}-s^{2}-24=0 $$ (hint: some answers may include i)

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Factor the expression by following steps:
- step0: Factor:
  $$x^{3}+x^{2}+4x+4$$
- step1: Calculate:
  $$x^{3}+4x+x^{2}+4$$
- step2: Rewrite the expression:
  $$x\times x^{2}+x\times 4+x^{2}+4$$
- step3: Factor the expression:
  $$x\left(x^{2}+4\right)+x^{2}+4$$
- step4: Factor the expression:
  $$\left(x+1\right)\left(x^{2}+4\right)$$
Solve the equation $$ n^{3}+5 n^{2}-9 n-45=0 $$.
Solve the equation by following steps:
- step0: Solve for $$n$$:
  $$n^{3}+5n^{2}-9n-45=0$$
- step1: Factor the expression:
  $$\left(n-3\right)\left(n+3\right)\left(n+5\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&n-3=0\&n+3=0\&n+5=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&n=3\&n=-3\&n=-5\end{align}$$
- step4: Rewrite:
  $$n_{1}=-5,n_{2}=-3,n_{3}=3$$
Solve the equation $$ 3 s^{4}-s^{2}-24=0 $$.
Solve the equation by following steps:
- step0: Solve for $$s$$:
  $$3s^{4}-s^{2}-24=0$$
- step1: Factor the expression:
  $$\left(s^{2}-3\right)\left(3s^{2}+8\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&s^{2}-3=0\&3s^{2}+8=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&s=\sqrt{3}\&s=-\sqrt{3}\&s \notin \mathbb{R}\end{align}$$
- step4: Find the union:
  $$\begin{align}&s=\sqrt{3}\&s=-\sqrt{3}\end{align}$$
- step5: Rewrite:
  $$s_{1}=-\sqrt{3},s_{2}=\sqrt{3}$$
Factor the expression $$ x^{4}-25 $$.
Factor the expression by following steps:
- step0: Factor:
  $$x^{4}-25$$
- step1: Rewrite the expression:
  $$\left(x^{2}\right)^{2}-5^{2}$$
- step2: Factor the expression:
  $$\left(x^{2}-5\right)\left(x^{2}+5\right)$$
Solve the equation $$ y^{3}-7 y^{2}-4 y+28=0 $$.
Solve the equation by following steps:
- step0: Solve for $$y$$:
  $$y^{3}-7y^{2}-4y+28=0$$
- step1: Factor the expression:
  $$\left(y-7\right)\left(y-2\right)\left(y+2\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&y-7=0\&y-2=0\&y+2=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&y=7\&y=2\&y=-2\end{align}$$
- step4: Rewrite:
  $$y_{1}=-2,y_{2}=2,y_{3}=7$$
Factor the expression $$ a^{4}+5 a^{2}-6 $$.
Factor the expression by following steps:
- step0: Factor:
  $$a^{4}+5a^{2}-6$$
- step1: Rewrite the expression:
  $$a^{4}+\left(6-1\right)a^{2}-6$$
- step2: Calculate:
  $$a^{4}+6a^{2}-a^{2}-6$$
- step3: Rewrite the expression:
  $$a^{2}\times a^{2}+a^{2}\times 6-a^{2}-6$$
- step4: Factor the expression:
  $$a^{2}\left(a^{2}+6\right)-\left(a^{2}+6\right)$$
- step5: Factor the expression:
  $$\left(a^{2}-1\right)\left(a^{2}+6\right)$$
- step6: Factor the expression:
  $$\left(a-1\right)\left(a+1\right)\left(a^{2}+6\right)$$
Let's go through each problem step by step:

1) **Factor the polynomial completely: $$ x^{3}+x^{2}+4x+4 $$**

To factor the polynomial, we can group the terms:
   $$
   x^{3}+x^{2}+4x+4 = (x^{3}+x^{2}) + (4x+4) = x^{2}(x+1) + 4(x+1) = (x+1)(x^{2}+4)
   $$
   Thus, the factored form is:
   $$
   (x+1)(x^{2}+4)
   $$

2) **Solve: $$ y^{3}-7y^{2}-4y+28=0 $$**

Using the Rational Root Theorem and synthetic division, we find the roots:
   - The roots are $$ y_{1} = -2 $$, $$ y_{2} = 2 $$, and $$ y_{3} = 7 $$.

3) **Solve: $$ n^{3}+5n^{2}-9n=45 $$**

Rearranging gives us:
   $$
   n^{3}+5n^{2}-9n-45=0
   $$
   Using synthetic division or factoring, we find the roots:
   - The roots are $$ n_{1} = -5 $$, $$ n_{2} = -3 $$, and $$ n_{3} = 3 $$.

4) **Factor the polynomial completely: $$ x^{4}-25 $$**

This is a difference of squares:
   $$
   x^{4}-25 = (x^{2}-5)(x^{2}+5)
   $$

5) **Factor the polynomial completely: $$ a^{4}+5a^{2}-6 $$**

Let $$ b = a^{2} $$, then we have:
   $$
   b^{2}+5b-6 = (b-1)(b+6)
   $$
   Substituting back $$ b = a^{2} $$:
   $$
   (a^{2}-1)(a^{2}+6) = (a-1)(a+1)(a^{2}+6)
   $$

6) **Solve: $$ 3s^{4}-s^{2}-24=0 $$**

Let $$ u = s^{2} $$, then we have:
   $$
   3u^{2}-u-24=0
   $$
   Using the quadratic formula:
   $$
   u = \frac{-(-1) \pm \sqrt{(-1)^{2}-4 \cdot 3 \cdot (-24)}}{2 \cdot 3} = \frac{1 \pm \sqrt{1+288}}{6} = \frac{1 \pm 17}{6}
   $$
   This gives us $$ u_{1} = 3 $$ and $$ u_{2} = -4 $$. Since $$ u = s^{2} $$, we have:
   - $$ s_{1} = -\sqrt{3} $$
   - $$ s_{2} = \sqrt{3} $$

In summary, the solutions and factorizations are:
1. $$ (x+1)(x^{2}+4) $$
2. Roots: $$ y = -2, 2, 7 $$
3. Roots: $$ n = -5, -3, 3 $$
4. $$ (x^{2}-5)(x^{2}+5) $$
5. $$ (a-1)(a+1)(a^{2}+6) $$
6. Roots: $$ s = -\sqrt{3}, \sqrt{3} $$
1. $$ (x+1)(x^{2}+4) $$ 2. $$ y = -2, 2, 7 $$ 3. $$ n = -5, -3, 3 $$ 4. $$ (x^{2}-5)(x^{2}+5) $$ 5. $$ (a-1)(a+1)(a^{2}+6) $$ 6. $$ s = -\sqrt{3}, \sqrt{3} $$

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