Q. \(x^2 + 2x – 3\)

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

\Delta=2^2-4(1)(-3)=16. Roots: x=\frac{-2\pm4}{2}, so x=1 or x=-3.

Q: Factor x^2+2x-3.

Look for (x+a)(x+b) with ab=-3 and a+b=2. Thus x^2+2x-3=(x+3)(x-1).

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

x=\frac{-2\pm\sqrt{4+12}}{2}=\frac{-2\pm4}{2}. So x=1 or x=-3.

Q: Complete the square for x^2+2x-3.

x^2+2x=(x+1)^2-1. So x^2+2x-3=(x+1)^2-4.

Q: What is the vertex of y=x^2+2x-3?

For y=ax^2+bx+c, vertex x=-\frac{b}{2a}=-\frac{2}{2}=-1. y=(-1)^2+2(-1)-3=-4. Vertex (-1,-4).

Q: Determine the y-intercept of x^2+2x-3.

Set x=0: y=-3. So the y-intercept is (0,-3).

Q: Compute the discriminant of x^2+2x-3.

\Delta=b^2-4ac=2^2-4(1)(-3)=4+12=16. Since \Delta>0, there are two real roots.

Answer

\(x^2+2x-3\) factors by finding numbers that multiply to \(-3\) and add to \(2\): \(3\) and \(-1\).

\[
x^2+2x-3=(x+3)(x-1)
\]

So the factored form is \((x+3)(x-1)\). If you need the zeros, set each factor to \(0\): \(x=-3\) or \(x=1\).

Detailed Explanation

We want to simplify (and typically factor) the expression \(x^2 + 2x – 3\).

Step 1: Identify the expression

The expression is:

\[
x^2 + 2x – 3
\]

Step 2: Factor the quadratic (if possible)

For a quadratic of the form \(ax^2 + bx + c\), we look for two numbers that multiply to \(ac\) and add to \(b\).

Here, \(a = 1\), \(b = 2\), and \(c = -3\).

So we need two numbers such that:

\[
(\text{number 1})(\text{number 2}) = (1)(-3) = -3
\]

and

\[
(\text{number 1}) + (\text{number 2}) = 2
\]

Step 3: Find the pair of numbers

The factors of \(-3\) are \(3\) and \(-1\). Their product is \(-3\), and their sum is \(3 + (-1) = 2\).

So we match:

\[
x^2 + 2x – 3 = (x + 3)(x – 1)
\]

Final Answer

\[
\boxed{(x + 3)(x – 1)}
\]

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Algebra FAQ

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

\(\Delta=2^2-4(1)(-3)=16\). Roots: \(x=\frac{-2\pm4}{2}\), so \(x=1\) or \(x=-3\).

Factor \(x^2+2x-3\).

Look for \((x+a)(x+b)\) with \(ab=-3\) and \(a+b=2\). Thus \(x^2+2x-3=(x+3)(x-1)\).

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

\(x=\frac{-2\pm\sqrt{4+12}}{2}=\frac{-2\pm4}{2}\). So \(x=1\) or \(x=-3\).

Complete the square for \(x^2+2x-3\).

\(x^2+2x=(x+1)^2-1\). So \(x^2+2x-3=(x+1)^2-4\).

What is the vertex of \(y=x^2+2x-3\)?

For \(y=ax^2+bx+c\), vertex \(x=-\frac{b}{2a}=-\frac{2}{2}=-1\). \(y=(-1)^2+2(-1)-3=-4\). Vertex \((-1,-4)\).

Determine the y-intercept of \(x^2+2x-3\).

Set \(x=0\): \(y=-3\). So the y-intercept is \((0,-3)\).

Compute the discriminant of \(x^2+2x-3\).

\(\Delta=b^2-4ac=2^2-4(1)(-3)=4+12=16\). Since \(\Delta>0\), there are two real roots.

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