Q. \(x^2+12x+c\)

Q: What values of c make x^2+12x+c factor over integers?

Use x^2+12x+c=(x+6)^2+(c-36). For integer factors: c-36=ab where (x+6)^2 shifts to (x+m)(x+n) with m+n=12 and mn=c.

Q: What are the roots in terms of c?

Solve \,x^2+12x+c=0. \[ x=\frac{-12\pm\sqrt{144-4c}}{2}=-6\pm\sqrt{36-c}. \]

Q: When are the roots real?

Need discriminant \,144-4c\ge 0\Rightarrow 36-c\ge 0\Rightarrow c\le 36.

Q: For what c does the quadratic have a double root?

Double root means discriminant 0: 144-4c=0\Rightarrow c=36. Then x=-6.

Q: What c makes the quadratic always positive?

Since minimum is c-36, always positive means c-36>0\Rightarrow c>36. Always nonnegative means c\ge 36.

Q: For what c does the quadratic cross the x-axis?

It crosses if real roots exist: c\le 36. It touches at one point if c=36. It doesn’t cross if c>36.

Answer

Let \(x^2+12x+c\) be rewritten by completing the square.

\[
x^2+12x+c=(x+6)^2-36+c=(x+6)^2+(c-36).
\]

So the simplified completed-square form is \((x+6)^2+(c-36)\).

Detailed Explanation

Problem: Simplify the expression \(x^2 + 12x + c\) by writing it in a fully factored (or completed-square) form.

Step 1: Identify the form to use. The expression is quadratic in \(x\):

\[
x^2 + 12x + c
\]

To rewrite a quadratic in a structured way, two common methods are:

Factoring (if possible).
Completing the square (always works for quadratics).

Because the constant term is \(c\) (unknown), factoring may require assumptions about \(c\). So the guaranteed method is completing the square.

Step 2: Group the terms involving \(x\).

\[
x^2 + 12x + c = \left(x^2 + 12x\right) + c
\]

Step 3: Complete the square for \(x^2 + 12x\).

We want to express \(x^2 + 12x\) in the form \(\left(x + a\right)^2\). Recall:

\[
\left(x + a\right)^2 = x^2 + 2ax + a^2
\]

In \(x^2 + 12x\), the coefficient of \(x\) is \(12\). So we match:

\[
2a = 12
\]

Therefore:

\[
a = 6
\]

Step 4: Add and subtract the needed constant.

Compute \(a^2\):

\[
a^2 = 6^2 = 36
\]

Now rewrite \(x^2 + 12x\) by adding and subtracting \(36\):

\[
x^2 + 12x = x^2 + 12x + 36 – 36
\]

Step 5: Convert the perfect square.

\[
\left(x^2 + 12x + 36\right) – 36 = \left(x + 6\right)^2 – 36
\]

Step 6: Substitute back into the original expression.

We had:

\[
\left(x^2 + 12x\right) + c
\]

So replace \(x^2 + 12x\) with \(\left(x + 6\right)^2 – 36\):

\[
x^2 + 12x + c = \left(\left(x + 6\right)^2 – 36\right) + c
\]

Step 7: Simplify constants.

\[
x^2 + 12x + c = \left(x + 6\right)^2 + c – 36
\]

Reorder the constants:

\[
x^2 + 12x + c = \left(x + 6\right)^2 + \left(c – 36\right)
\]

Final Answer (completed-square form):

\[
x^2 + 12x + c = \left(x + 6\right)^2 + \left(c – 36\right)
\]

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Algebra FAQ

What values of \(c\) make \(x^2+12x+c\) factor over integers?

Use \(x^2+12x+c=(x+6)^2+(c-36)\). For integer factors: \(c-36=ab\) where \((x+6)^2\) shifts to \((x+m)(x+n)\) with \(m+n=12\) and \(mn=c\).

What are the roots in terms of \(c\)?

Solve \(\,x^2+12x+c=0\). \[ x=\frac{-12\pm\sqrt{144-4c}}{2}=-6\pm\sqrt{36-c}. \]

When are the roots real?

Need discriminant \(\,144-4c\ge 0\Rightarrow 36-c\ge 0\Rightarrow c\le 36\).

What is the vertex and minimum value?

Complete the square: \[ x^2+12x+c=(x+6)^2+(c-36). \] Vertex is \((-6,c-36)\). Minimum value is \(c-36\).

For what \(c\) does the quadratic have a double root?

Double root means discriminant \(0\): \(144-4c=0\Rightarrow c=36\). Then \(x=-6\).

What \(c\) makes the quadratic always positive?

Since minimum is \(c-36\), always positive means \(c-36>0\Rightarrow c>36\). Always nonnegative means \(c\ge 36\).

For what \(c\) does the quadratic cross the \(x\)-axis?

It crosses if real roots exist: \(c\le 36\). It touches at one point if \(c=36\). It doesn’t cross if \(c>36\).

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