Q. Solve for all values of \(x\) by factoring. Start with the equation \(x^2 – 5x – 1 = -1\).

Q: What is the first step to solve x^2-5x-1=-1 by factoring?

Add 1 to both sides to get all terms on one side: x^2-5x-1+1=0, which simplifies to x^2-5x=0. Factoring requires a zero on one side..

Q: How do you factor x^2-5x?.

Factor out the greatest common factor x: x^2-5x=x(x-5).

Q: How do you get the solutions from x(x-5)=0?.

Use the zero product property: if ab=0, then a=0 or b=0. So x=0 or x-5=0, which implies x=5.

Q: How can I check the solutions are correct?

Substitute into the original equation x^2-5x-1=-1. For x=0: 0-0-1=-1. For x=5: 25-25-1=-1. Both satisfy the equation.

Q: What if the quadratic didn't factor nicely?

Use the quadratic formula x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} or complete the square. These work for any quadratic, even when factoring over integers fails.

Q: Why is it valid to factor out x here?

Why is it valid to factor out x here?

Q: How would the process change if the constant were different, e.g., x^2-5x-6=-1?.

First move constants: x^2-5x-6+1=0\Rightarrow x^2-5x-5=0. If it doesn't factor nicely, use the quadratic formula or complete the square.

Answer

\[
x^2-5x-1=-1
\]
Add 1 to both sides:
\[
x^2-5x=0
\]
Factor:
\[
x(x-5)=0
\]
Thus \(x=0\) or \(x=5\).

Detailed Explanation

Given equation

\(x^2 – 5x – 1 = -1\)

Step 1 — Isolate the polynomial on one side

Add 1 to both sides to eliminate the constant term on the left:
\(x^2 – 5x – 1 + 1 = -1 + 1\).
Simplify each side:
\(x^2 – 5x = 0\).

Step 2 — Factor the left-hand side

Factor out the greatest common factor, which is \(x\):
\(x(x – 5) = 0\).

Step 3 — Apply the zero-product property

If a product of two factors equals zero, then at least one factor must be zero. So set each factor equal to zero separately:
\(x = 0\) or \(x – 5 = 0\).

Step 4 — Solve each simple equation

From \(x = 0\) we have the solution \(x = 0\).

From \(x – 5 = 0\) add 5 to both sides to get \(x = 5\).

Final answer

The solutions are \(x = 0\) and \(x = 5\).

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Algebra FAQs

What is the first step to solve \(x^2-5x-1=-1\) by factoring?

Add 1 to both sides to get all terms on one side: \(x^2-5x-1+1=0\), which simplifies to \(x^2-5x=0\). Factoring requires a zero on one side..

How do you factor \(x^2-5x\)?.

Factor out the greatest common factor \(x\): \(x^2-5x=x(x-5)\).

How do you get the solutions from \(x(x-5)=0\)?.

Use the zero product property: if \(ab=0\), then \(a=0\) or \(b=0\). So \(x=0\) or \(x-5=0\), which implies \(x=5\).

How can I check the solutions are correct?

Substitute into the original equation \(x^2-5x-1=-1\). For \(x=0\): \(0-0-1=-1\). For \(x=5\): \(25-25-1=-1\). Both satisfy the equation.

What if the quadratic didn't factor nicely?

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) or complete the square. These work for any quadratic, even when factoring over integers fails.

Why is it valid to factor out \(x\) here?

Could there be extraneous solutions from factoring?

No. Factoring and applying the zero product property preserves equivalence when you start from an equation set equal to zero, so all solutions found are valid.

How would the process change if the constant were different, e.g., \(x^2-5x-6=-1\)?.

First move constants: \(x^2-5x-6+1=0\Rightarrow x^2-5x-5=0\). If it doesn't factor nicely, use the quadratic formula or complete the square.

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