Q. \(x^2-4x-5\)

Q: Factor x^2-4x-5.

Find two numbers with sum -4 and product -5: -5 and 1. So x^2-4x-5=(x-5)(x+1).

Q: Solve x^2-4x-5=0.

Using factors (x-5)(x+1)=0. So x=5 or x=-1.

Q: Use the quadratic formula for x^2-4x-5=0.

x=\dfrac{4\pm\sqrt{16+20}}{2}=\dfrac{4\pm 6}{2}, so x=5 or x=-1.

Q: What are the x-intercepts of y=x^2-4x-5?

Set y=0: (x-5)(x+1)=0. Intercepts are at x=5 and x=-1, so points (5,0) and (-1,0).

Q: What are the vertex and minimum value of y=x^2-4x-5?

Vertex at x=\dfrac{-(-4)}{2}=2. Value: y=(2-2)^2-9=-9. Minimum is -9 since it opens up.

Answer

To factor the quadratic \(x^2-4x-5\), find two numbers that multiply to \(-5\) and add to \(-4\). They are \(-5\) and \(+1\).

\[
x^2-4x-5=(x-5)(x+1)
\]

Final answer: \((x-5)(x+1)\).

Detailed Explanation

We want to factor and simplify the expression \(x^2 – 4x – 5\).

Step 1: Identify the types of terms

The expression is a quadratic in the form \(ax^2 + bx + c\), where:

\(a = 1\), \(b = -4\), \(c = -5\).

Step 2: Factor using the \(ac\) method

First compute the product \(ac\):

\[
ac = 1 \cdot (-5) = -5
\]

Now we need two numbers that:

Multiply to \(-5\)
Add to \(-4\)

The numbers are \(-5\) and \(1\), because:

\(-5 \cdot 1 = -5\)
\(-5 + 1 = -4\)

Step 3: Rewrite the middle term

Replace \(-4x\) with \(-5x + x\):

\[
x^2 – 4x – 5 = x^2 – 5x + x – 5
\]

Step 4: Factor by grouping

Group the terms into two pairs:

\[
x^2 – 5x + x – 5 = (x^2 – 5x) + (x – 5)
\]

Factor each group:

\[
(x^2 – 5x) = x(x – 5)
\]
\[
(x – 5) = 1(x – 5)
\]

So the expression becomes:

\[
x(x – 5) + 1(x – 5)
\]

Step 5: Factor out the common binomial

Both terms contain \((x – 5)\), so factor it out:

\[
x(x – 5) + 1(x – 5) = (x + 1)(x – 5)
\]

Final Answer

\[
x^2 – 4x – 5 = (x + 1)(x – 5)
\]

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Algebra FAQ

Factor \(x^2-4x-5\).

Find two numbers with sum \(-4\) and product \(-5\): \(-5\) and \(1\). So \(x^2-4x-5=(x-5)(x+1)\).

Solve \(x^2-4x-5=0\).

Using factors \((x-5)(x+1)=0\). So \(x=5\) or \(x=-1\).

What is the completed square form of \(x^2-4x-5\)?

\(x^2-4x-5=(x-2)^2-9\).

Use the quadratic formula for \(x^2-4x-5=0\).

\(x=\dfrac{4\pm\sqrt{16+20}}{2}=\dfrac{4\pm 6}{2}\), so \(x=5\) or \(x=-1\).

What are the \(x\)-intercepts of \(y=x^2-4x-5\)?

Set \(y=0\): \((x-5)(x+1)=0\). Intercepts are at \(x=5\) and \(x=-1\), so points \((5,0)\) and \((-1,0)\).

What are the vertex and minimum value of \(y=x^2-4x-5\)?

Vertex at \(x=\dfrac{-(-4)}{2}=2\). Value: \(y=(2-2)^2-9=-9\). Minimum is \(-9\) since it opens up.

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