Q. \(x^{2}+3x-10\)

Q: Find the roots of x^2+3x-10=0.

Roots are x=-5 and x=2, since (x+5)(x-2)=0.

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

x-intercepts from the zeros: (-5,0) and (2,0).

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

For y=ax^2+bx+c, x_v=-\frac{b}{2a}=-\frac{3}{2}. Then y_v=-\frac{25}{4}-10=-\frac{65}{4}.

Q: What is the value at x=0 (the y-intercept)?

f(0)=0+0-10=-10, so the y-intercept is (0,-10).

Answer

We factor the quadratic by finding two numbers that multiply to \(-10\) and add to \(3\): \(5\) and \(-2\).

\[
x^2+3x-10=(x+5)(x-2)
\]

So the expression factors as \((x+5)(x-2)\).

Detailed Explanation

We want to factor the expression \(x^2+3x-10\). Factoring a quadratic means writing it as a product of two simpler expressions.

Step 1: Identify the quadratic form

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

\(a=1\)
\(b=3\)
\(c=-10\)

Step 2: Find two numbers whose product is \(ac\)

Compute the product \(ac\):

\[
ac = 1 \cdot (-10) = -10
\]

So we need two integers that multiply to \(-10\).

The possible pairs are:

\(1\) and \(-10\)
\(-1\) and \(10\)
\(2\) and \(-5\)
\(-2\) and \(5\)

Step 3: Make their sum equal to \(b\)

We also need the sum of the same two numbers to equal \(b=3\).

Check each pair:

\(1+(-10)=-9\) (not \(3\))
\((-1)+10=9\) (not \(3\))
\(2+(-5)=-3\) (not \(3\))
\((-2)+5=3\) (this works)

Step 4: Write the factorization

The numbers are \(-2\) and \(5\). For a trinomial \(x^2+3x-10\), the factorization is:

\[
x^2+3x-10 = (x-2)(x+5)
\]

Final Answer

\[
x^2+3x-10 = (x-2)(x+5)
\]

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Algebra FAQ

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

\(x^2+3x-10=(x+5)(x-2)\).

Find the roots of \(x^2+3x-10=0\).

Roots are \(x=-5\) and \(x=2\), since \((x+5)(x-2)=0\).

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

\(x=\frac{-3\pm\sqrt{3^2-4(1)(-10)}}{2}=\frac{-3\pm\sqrt{49}}{2}=\frac{-3\pm 7}{2}\).

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

\(x\)-intercepts from the zeros: \((-5,0)\) and \((2,0)\).

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

For \(y=ax^2+bx+c\), \(x_v=-\frac{b}{2a}=-\frac{3}{2}\). Then \(y_v=-\frac{25}{4}-10=-\frac{65}{4}\).

What is the value at \(x=0\) (the \(y\)-intercept)?

\(f(0)=0+0-10=-10\), so the \(y\)-intercept is \((0,-10)\).

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