Q. \[ x^2 + 5x + 7 \]

Q: Can x^2+5x+7 be factored over integers?

Discriminant is 25-28=-3<0, so it cannot factor into real linear factors with integer coefficients.

Q: What are the roots of x^2+5x+7=0?

Using x=\frac{-5\pm\sqrt{-3}}{2}, roots are x=\frac{-5\pm i\sqrt{3}}{2}.

Q: What is the completed square form?

x^2+5x+7=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}+7=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}+\frac{28}{4}=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}.

Q: What is the vertex and minimum value of x^2+5x+7?

Vertex at x=-\frac{5}{2}. Minimum value is f\left(-\frac{5}{2}\right)=\frac{3}{4}.

Q: Is x^2+5x+7 always positive?

Yes, since it equals \left(x+\frac{5}{2}\right)^2+\frac{3}{4}, and \left(x+\frac{5}{2}\right)^2\ge 0, so the expression is \ge \frac{3}{4}.

Q: What is the axis of symmetry?

The parabola x^2+5x+7 is symmetric about x=-\frac{5}{2}.

Answer

To factor \(x^2+5x+7\), check numbers that multiply to \(7\) and add to \(5\). No integers satisfy this, so it does not factor nicely over the integers. Its roots come from the quadratic formula.

\[
x^2+5x+7=0
\]
\[
x=\frac{-5\pm\sqrt{5^2-4\cdot1\cdot7}}{2}=\frac{-5\pm\sqrt{25-28}}{2}=\frac{-5\pm\sqrt{-3}}{2}
\]

Final result:

\[
x=\frac{-5\pm i\sqrt{3}}{2}
\]

Detailed Explanation

We are asked to work with the expression \(x^2 + 5x + 7\). This is a quadratic polynomial.

Step 1: Identify the coefficients.

A quadratic has the form \(ax^2 + bx + c\). Here:

\(a = 1\)
\(b = 5\)
\(c = 7\)

Step 2: Check whether it can be factored over the integers.

To factor \(x^2 + 5x + 7\) (into integer factors), we would look for numbers that multiply to \(7\) and add to \(5\).

The factor pairs of \(7\) are:

\(1\) and \(7\) (but \(1 + 7 = 8 \ne 5\))
\(-1\) and \(-7\) (but \(-1 + (-7) = -8 \ne 5\))

So, there are no integer numbers that work.

Step 3: Conclude the factoring form (no integer factorization).

Because no such pair exists, \(x^2 + 5x + 7\) cannot be factored using integer coefficients. It remains in expanded polynomial form.

Final answer:

\(x^2 + 5x + 7\)

See full solution

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Algebra FAQ

Can \(x^2+5x+7\) be factored over integers?

Discriminant is \(25-28=-3<0\), so it cannot factor into real linear factors with integer coefficients.

What are the roots of \(x^2+5x+7=0\)?

Using \(x=\frac{-5\pm\sqrt{-3}}{2}\), roots are \(x=\frac{-5\pm i\sqrt{3}}{2}\).

What is the completed square form?

\(x^2+5x+7=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}+7=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}+\frac{28}{4}=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\).

What is the vertex and minimum value of \(x^2+5x+7\)?

Vertex at \(x=-\frac{5}{2}\). Minimum value is \(f\left(-\frac{5}{2}\right)=\frac{3}{4}\).

Is \(x^2+5x+7\) always positive?

Yes, since it equals \(\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\), and \(\left(x+\frac{5}{2}\right)^2\ge 0\), so the expression is \(\ge \frac{3}{4}\).

What is the axis of symmetry?

The parabola \(x^2+5x+7\) is symmetric about \(x=-\frac{5}{2}\).

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