Q. \(x^2 – 4x + 24\)

Q: What are the roots of x^2-4x+24=0?

Compute discriminant b^2-4ac=16-96=-80<0, so no real roots. Complex roots: x=\frac{4\pm\sqrt{-80}}{2}=2\pm 2\sqrt{5}\,i.

Q: Can you factor x^2-4x+24 over real numbers?

No. The discriminant is negative -80, so it cannot factor into real linear factors like (x-r)(x-s).

Q: What is the vertex form of x^2-4x+24?

Complete the square: x^2-4x+24=(x-2)^2+20.

Q: What is the minimum value of x^2-4x+24 for real x?

Since (x-2)^2\ge 0, the minimum is 20, achieved at x=2.

Q: Is the quadratic always positive or can it be negative?

Always positive because x^2-4x+24=(x-2)^2+20\ge 20>0. So it never goes negative and has no real zeros.

Q: What are the axis of symmetry and direction of opening?

Axis of symmetry is x=2. Since the leading coefficient is 1>0, it opens upward.

Answer

We complete the square by factoring the quadratic:

\[
x^2 – 4x + 24 = (x-2)^2 + 20
\]

This quadratic has discriminant

\[
\Delta = (-4)^2 – 4(1)(24) = 16 – 96 = -80 < 0
\]

So it has no real zeros and is always positive.

Final result: \((x-2)^2 + 20\)

Detailed Explanation

We are given the expression

\[
x^2 – 4x + 24
\]

Since the problem is just to write out or simplify the expression, we first check whether it can be factored into smaller polynomial factors.

The expression is a quadratic of the form

\[
ax^2 + bx + c
\]

Here,

\[
a = 1,\quad b = -4,\quad c = 24
\]

To factor \(x^2 – 4x + 24\), we would need two numbers that:

1. Multiply to \(24\), but

2. Add to \(-4\).

List factor pairs of \(24\):

\[
1\cdot 24 = 24,\quad 2\cdot 12 = 24,\quad 3\cdot 8 = 24,\quad 4\cdot 6 = 24
\]

Now check their sums:

\(1 + 24 = 25\) does not equal \(-4\).

\(2 + 12 = 14\) does not equal \(-4\).

\(3 + 8 = 11\) does not equal \(-4\).

\(4 + 6 = 10\) does not equal \(-4\).

Try negative possibilities that would give a negative sum:

\(-1 + (-24) = -25\) does not equal \(-4\).

\(-2 + (-12) = -14\) does not equal \(-4\).

\(-3 + (-8) = -11\) does not equal \(-4\).

\(-4 + (-6) = -10\) does not equal \(-4\).

So there are no integers that satisfy both conditions, meaning the quadratic does not factor nicely over the integers.

To further confirm, we can compute the discriminant \( \Delta \) of the quadratic.

\[
\Delta = b^2 – 4ac
\]

Substitute \(a=1\), \(b=-4\), \(c=24\):

\[
\Delta = (-4)^2 – 4(1)(24)
\]

\[
\Delta = 16 – 96
\]

\[
\Delta = -80
\]

Because the discriminant is negative (\(-80\)), the quadratic has no real (and no integer) factorization.

Therefore, the expression is already in simplest polynomial form.

Final answer:

\[
x^2 – 4x + 24
\]

See full solution

Graph

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Algebra FAQ

What are the roots of \(x^2-4x+24=0\)?

Compute discriminant \(b^2-4ac=16-96=-80<0\), so no real roots. Complex roots: \(x=\frac{4\pm\sqrt{-80}}{2}=2\pm 2\sqrt{5}\,i\).

Can you factor \(x^2-4x+24\) over real numbers?

No. The discriminant is negative \(-80\), so it cannot factor into real linear factors like \((x-r)(x-s)\).

What is the vertex form of \(x^2-4x+24\)?

Complete the square: \(x^2-4x+24=(x-2)^2+20\).

What is the minimum value of \(x^2-4x+24\) for real \(x\)?

Since \((x-2)^2\ge 0\), the minimum is \(20\), achieved at \(x=2\).

Is the quadratic always positive or can it be negative?

Always positive because \(x^2-4x+24=(x-2)^2+20\ge 20>0\). So it never goes negative and has no real zeros.

What are the axis of symmetry and direction of opening?

Axis of symmetry is \(x=2\). Since the leading coefficient is \(1>0\), it opens upward.

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