Q. \(x^2 – 2x – 15\)

Q: . Factor x^2-2x-15 completely?

. Find two numbers with sum -2 and product -15: -5 and 3. So x^2-2x-15=(x-5)(x+3).

Q: . Solve x^2-2x-15=0?

. Use the factorization: (x-5)(x+3)=0. Thus x=5 or x=-3.

Q: . What are the x-intercepts and y-intercept?

. x-intercepts: (5,0) and (-3,0). For y-intercept, plug in x=0: y=-15, so (0,-15).

Q: . Find the vertex of the parabola y=x^2-2x-15?

. Vertex x-value is -\frac{b}{2a}=\frac{2}{2}=1. Then y(1)=1-2-15=-16. Vertex: (1,-16).

Q: . What is the axis of symmetry?

. For y=ax^2+bx+c, axis is x=-\frac{b}{2a}=\frac{2}{2}=1. So the axis is x=1.

Q: . Can you solve using the quadratic formula?

. For a=1, b=-2, c=-15: x=\frac{2\pm\sqrt{(-2)^2-4(1)(-15)}}{2}=\frac{2\pm\sqrt{64}}{2}=5,-3.

Answer

To factor \(x^2-2x-15\), find two numbers that multiply to \(-15\) and add to \(-2\). Those numbers are \(-5\) and \(3\).

\[
x^2-2x-15=(x-5)(x+3)
\]

Final result: \((x-5)(x+3)\).

Detailed Explanation

We want to work with the expression \(x^2 – 2x – 15\). A common goal is to factor it, if possible.

Step 1: Write the quadratic in standard form

The expression is already a quadratic in the form

\[
ax^2 + bx + c
\]

Here, \(a = 1\), \(b = -2\), and \(c = -15\).

Step 2: Find two numbers that multiply to \(ac\)

Compute \(ac\):

\[
ac = 1 \cdot (-15) = -15
\]

So we need two numbers that multiply to \(-15\).

Step 3: Find two numbers that add to \(b\)

We need two numbers whose sum is \(b = -2\).

List factor pairs of \(-15\):

\[
-1 \text{ and } 15 \quad (\text{sum } 14)
\]

\[
1 \text{ and } -15 \quad (\text{sum } -14)
\]

\[
3 \text{ and } -5 \quad (\text{sum } -2)
\]

\[
-3 \text{ and } 5 \quad (\text{sum } 2)
\]

The pair that works is \(3\) and \(-5\) because their product is \(-15\) and their sum is \(-2\).

Step 4: Factor the quadratic using the found numbers

Since the numbers are \(3\) and \(-5\), we factor as follows:

\[
x^2 – 2x – 15 = (x + 3)(x – 5)
\]

Explanation: The product \((x + 3)(x – 5)\) gives \(x^2\), and the middle terms combine to \(-2x\), while the constant term is \(-15\).

Final Answer

\[
x^2 – 2x – 15 = (x + 3)(x – 5)
\]

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Algebra FAQ

. Factor \(x^2-2x-15\) completely?

. Find two numbers with sum \(-2\) and product \(-15\): \(-5\) and \(3\). So \(x^2-2x-15=(x-5)(x+3)\).

. Solve \(x^2-2x-15=0\)?

. Use the factorization: \((x-5)(x+3)=0\). Thus \(x=5\) or \(x=-3\).

. What are the zeros of \(x^2-2x-15\)?

. Set the expression to zero to get the roots: zeros occur at \(x=5\) and \(x=-3\).

. What are the \(x\)-intercepts and \(y\)-intercept?

. \(x\)-intercepts: \((5,0)\) and \((-3,0)\). For \(y\)-intercept, plug in \(x=0\): \(y=-15\), so \((0,-15)\).

. Find the vertex of the parabola \(y=x^2-2x-15\)?

. Vertex \(x\)-value is \(-\frac{b}{2a}=\frac{2}{2}=1\). Then \(y(1)=1-2-15=-16\). Vertex: \((1,-16)\).

. What is the axis of symmetry?

. For \(y=ax^2+bx+c\), axis is \(x=-\frac{b}{2a}=\frac{2}{2}=1\). So the axis is \(x=1\).

. Can you solve using the quadratic formula?

. For \(a=1\), \(b=-2\), \(c=-15\): \(x=\frac{2\pm\sqrt{(-2)^2-4(1)(-15)}}{2}=\frac{2\pm\sqrt{64}}{2}=5,-3\).

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