Question

1. ( x^2 - 7x + 6 )\\

factored form: \\
solution(s):

Explanation:

Step1: Try rational roots

We use the Rational Root Theorem. Possible rational roots are factors of 6 over factors of 1, so ±1, ±2, ±3, ±6. Test \(x = 1\): \(1^3 - 7(1)+6=1 - 7 + 6 = 0\). So \((x - 1)\) is a factor.

Step2: Perform polynomial division or use synthetic division

Using synthetic division with root 1:
\[

$$\begin{array}{r|rrr} 1 & 1 & 0 & -7 & 6\\ & & 1 & 1 & -6\\ \hline & 1 & 1 & -6 & 0 \end{array}$$

\]
So the polynomial factors as \((x - 1)(x^2+x - 6)\).

Step3: Factor the quadratic

Factor \(x^2+x - 6\). We need two numbers that multiply to -6 and add to 1. Those numbers are 3 and -2. So \(x^2+x - 6=(x + 3)(x - 2)\).

Step4: Combine factors

Thus, \(x^3 - 7x + 6=(x - 1)(x + 3)(x - 2)\).

Step5: Find solutions

Set each factor equal to zero:

• \(x - 1 = 0\) gives \(x = 1\)
• \(x + 3 = 0\) gives \(x=-3\)
• \(x - 2 = 0\) gives \(x = 2\)

Factored Form:

\((x - 1)(x + 3)(x - 2)\)

Solution(s):

\(x = 1\), \(x=-3\), \(x = 2\)

Answer:

Step1: Try rational roots

We use the Rational Root Theorem. Possible rational roots are factors of 6 over factors of 1, so ±1, ±2, ±3, ±6. Test \(x = 1\): \(1^3 - 7(1)+6=1 - 7 + 6 = 0\). So \((x - 1)\) is a factor.

Step2: Perform polynomial division or use synthetic division

Using synthetic division with root 1:
\[

$$\begin{array}{r|rrr} 1 & 1 & 0 & -7 & 6\\ & & 1 & 1 & -6\\ \hline & 1 & 1 & -6 & 0 \end{array}$$

\]
So the polynomial factors as \((x - 1)(x^2+x - 6)\).

Step3: Factor the quadratic

Factor \(x^2+x - 6\). We need two numbers that multiply to -6 and add to 1. Those numbers are 3 and -2. So \(x^2+x - 6=(x + 3)(x - 2)\).

Step4: Combine factors

Thus, \(x^3 - 7x + 6=(x - 1)(x + 3)(x - 2)\).

Step5: Find solutions

Set each factor equal to zero:

• \(x - 1 = 0\) gives \(x = 1\)
• \(x + 3 = 0\) gives \(x=-3\)
• \(x - 2 = 0\) gives \(x = 2\)

Factored Form:

\((x - 1)(x + 3)(x - 2)\)

Solution(s):

\(x = 1\), \(x=-3\), \(x = 2\)