Q. \(x^{2}-5x+6\)

Q: How do I factor x^{2}-5x+6?

Find two numbers that multiply to 6 and add to -5: -2 and -3. So x^{2}-5x+6=(x-2)(x-3).

Q: What are the roots of x^{2}-5x+6=0?

Set factored form to zero: (x-2)(x-3)=0. Thus x=2 or x=3.

Q: What is the discriminant of x^{2}-5x+6?

For ax^{2}+bx+c, \Delta=b^{2}-4ac. Here a=1, b=-5, c=6: \Delta=25-24=1.

Q: What is the vertex and minimum value of y=x^{2}-5x+6?

Vertex at x=-\frac{b}{2a}=\frac{5}{2}. Minimum value: y=\left(x-\frac{5}{2}\right)^{2}-\frac{1}{4}, so minimum is -\frac{1}{4}.

Q: How do I find the axis of symmetry?

For ax^{2}+bx+c, axis is x=-\frac{b}{2a}=\frac{5}{2}.

Q: What is the y-intercept of x^{2}-5x+6?

Plug in x=0: y=6. So the y-intercept is (0,6).

Answer

Factor the quadratic:

\[
x^{2}-5x+6
\]

We need two numbers that multiply to \(6\) and add to \(-5\). Those numbers are \(-2\) and \(-3\).

\[
x^{2}-5x+6=(x-2)(x-3)
\]

Detailed Explanation

We want to simplify the expression

\[
x^{2}-5x+6.
\]

This expression is a quadratic polynomial. A common first step is to factor it, if possible.

Step 1: Identify the quadratic form

A quadratic polynomial has the form

\[
ax^{2}+bx+c.
\]

Here,

\(a=1\)
\(b=-5\)
\(c=6\)

Step 2: Find two numbers that multiply to \(6\)

We need two integers \(m\) and \(n\) such that

\(m \cdot n = 6\)
\(m + n = -5\)

Let’s list factor pairs of \(6\):

\(1\) and \(6\) give sum \(7\)
\(-1\) and \(-6\) give sum \(-7\)
\(2\) and \(3\) give sum \(5\)
\(-2\) and \(-3\) give sum \(-5\)

The pair that works is \(-2\) and \(-3\), because

\[
(-2)(-3)=6
\]

and

\[
-2+(-3)=-5.
\]

Step 3: Factor the quadratic

Using those numbers, we can write

\[
x^{2}-5x+6=(x-2)(x-3).
\]

Final Answer

\[
x^{2}-5x+6=(x-2)(x-3).
\]

See full solution

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Algebra FAQ

How do I factor \(x^{2}-5x+6\)?

Find two numbers that multiply to \(6\) and add to \(-5\): \(-2\) and \(-3\). So \(x^{2}-5x+6=(x-2)(x-3)\).

What are the roots of \(x^{2}-5x+6=0\)?

Set factored form to zero: \((x-2)(x-3)=0\). Thus \(x=2\) or \(x=3\).

Can I complete the square for \(x^{2}-5x+6\)?

\(x^{2}-5x+6=\left(x-\frac{5}{2}\right)^{2}-\frac{25}{4}+6=\left(x-\frac{5}{2}\right)^{2}-\frac{1}{4}\).

What is the discriminant of \(x^{2}-5x+6\)?

For \(ax^{2}+bx+c\), \(\Delta=b^{2}-4ac\). Here \(a=1\), \(b=-5\), \(c=6\): \(\Delta=25-24=1\).

What is the vertex and minimum value of \(y=x^{2}-5x+6\)?

Vertex at \(x=-\frac{b}{2a}=\frac{5}{2}\). Minimum value: \(y=\left(x-\frac{5}{2}\right)^{2}-\frac{1}{4}\), so minimum is \(-\frac{1}{4}\).

How do I find the axis of symmetry?

For \(ax^{2}+bx+c\), axis is \(x=-\frac{b}{2a}=\frac{5}{2}\).

What is the y-intercept of \(x^{2}-5x+6\)?

Plug in \(x=0\): \(y=6\). So the y-intercept is \((0,6)\).

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