Q. \[ x^2 – 3x – 10 \]

Q: Find the roots of x^2-3x-10.

Factor: x^2-3x-10=(x-5)(x+2). Roots: x=5 or x=-2.

Q: Solve x^2-3x-10=0 using the quadratic formula.

a=1,b=-3,c=-10. x=\frac{3\pm\sqrt{9+40}}{2}=\frac{3\pm 7}{2}, so x=5 or x=-2.

Q: Factor x^2-3x-10 completely.

Seek numbers with product -10 and sum -3: -5 and 2. So x^2-3x-10=(x-5)(x+2).

Q: Determine the vertex of y=x^2-3x-10.

x-coordinate: x=\frac{-b}{2a}=\frac{3}{2}. Vertex y: \left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)-10=\frac{9}{4}-\frac{9}{2}-10=-\frac{49}{4}.

Q: What is the discriminant and what does it say about roots?

\Delta=b^2-4ac=9-4(1)(-10)=49. Since \Delta>0, there are two distinct real roots.

Answer

To factor \(x^2 – 3x – 10\), find two numbers with product \(-10\) and sum \(-3\). These are \(-5\) and \(2\).

\[
x^2 – 3x – 10 = (x – 5)(x + 2)
\]

Final result: \((x – 5)(x + 2)\)

Detailed Explanation

We want to factor the polynomial \(x^2 – 3x – 10\).

Step 1: Identify the polynomial in standard form.

The expression is already in standard quadratic form:

\[
x^2 – 3x – 10
\]
Here, \(a = 1\), \(b = -3\), and \(c = -10\).

Step 2: Find two numbers that multiply to \(c\) and add to \(b\).

We look for numbers \(m\) and \(n\) such that:

\[
m \cdot n = -10
\]
and
\[
m + n = -3
\]

Step 3: List factor pairs of \(-10\).

The factor pairs of \(-10\) are:

\[
1 \cdot (-10) = -10
\]
\[
(-1) \cdot 10 = -10
\]
\[
2 \cdot (-5) = -10
\]
\[
(-2) \cdot 5 = -10
\]

Step 4: Check which pair adds to \(-3\).

Compute sums:

\[
1 + (-10) = -9
\]
\[
(-1) + 10 = 9
\]
\[
2 + (-5) = -3
\]
\[
(-2) + 5 = 3
\]

The pair that works is \(2\) and \(-5\) because \(2 + (-5) = -3\).

Step 5: Use the factor pairs to factor the quadratic.

Since the numbers are \(2\) and \(-5\), we write:

\[
x^2 – 3x – 10 = (x + 2)(x – 5)
\]

Final answer:

\[
x^2 – 3x – 10 = (x + 2)(x – 5)
\]

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Algebra FAQ

Find the roots of \(x^2-3x-10\).

Factor: \(x^2-3x-10=(x-5)(x+2)\). Roots: \(x=5\) or \(x=-2\).

Solve \(x^2-3x-10=0\) using the quadratic formula.

\(a=1,b=-3,c=-10\). \(x=\frac{3\pm\sqrt{9+40}}{2}=\frac{3\pm 7}{2}\), so \(x=5\) or \(x=-2\).

Factor \(x^2-3x-10\) completely.

Seek numbers with product \(-10\) and sum \(-3\): \(-5\) and \(2\). So \(x^2-3x-10=(x-5)(x+2)\).

Determine the vertex of \(y=x^2-3x-10\).

\(x\)-coordinate: \(x=\frac{-b}{2a}=\frac{3}{2}\). Vertex \(y\): \( \left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)-10=\frac{9}{4}-\frac{9}{2}-10=-\frac{49}{4}\).

What is the discriminant and what does it say about roots?

\(\Delta=b^2-4ac=9-4(1)(-10)=49\). Since \(\Delta>0\), there are two distinct real roots.

Compute \(f(2)\) for \(f(x)=x^2-3x-10\).

\(f(2)=4-6-10=-12\).

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