Q. \(x^2 – 5x + 4\)

Q: Solve x^2-5x+4=0 by completing the square.

\left(x-\frac{5}{2}\right)^2=\frac{9}{4}, so x=\frac{5}{2}\pm\frac{3}{2}, giving 1,4.

Q: Compute the vertex of y=x^2-5x+4.

Vertex at x=\frac{-(-5)}{2\cdot 1}=\frac{5}{2}. y=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)+4=-\frac{1}{4}.

Q: Determine if x^2-5x+4 is always positive, always negative, or changes sign.

Discriminant 25-16=9>0 and leading coefficient is positive, so it changes sign at x=1 and x=4.

Answer

Factor the quadratic by finding two numbers that multiply to \(4\) and add to \(-5\). The numbers are \(-1\) and \(-4\).

\[
x^2-5x+4=(x-1)(x-4)
\]

Detailed Explanation

Problem: Solve (factor) the expression \(x^2 – 5x + 4\).

Step 1: Identify the quadratic form.

We have a quadratic polynomial in the form

\[
x^2 – 5x + 4
\]

Here, \(a = 1\), \(b = -5\), and \(c = 4\).

Step 2: Factor the quadratic.

To factor \(x^2 – 5x + 4\), we look for two numbers that:

Multiply to \(4\)
Add to \(-5\)

Step 3: Find the correct pair of numbers.

Consider the factor pairs of \(4\): \((1,4)\) and \((-1,-4)\).

Now check which pair adds to \(-5\):

\(1 + 4 = 5\)
\(-1 + (-4) = -5\)

The pair \(-1\) and \(-4\) works.

Step 4: Write the factored form.

Use these numbers to split the middle term \(-5x\) as \(-4x – x\):

\[
x^2 – 5x + 4 = x^2 – 4x – x + 4
\]

Now group the terms:

\[
x^2 – 4x – x + 4 = (x^2 – 4x) + (-x + 4)
\]

Factor each group:

\[
x^2 – 4x = x(x – 4)
\]

and

\[
-x + 4 = -(x – 4)
\]

So the expression becomes:

\[
(x(x – 4)) – (x – 4) = (x – 4)(x – 1)
\]

Therefore, the factorization is:

\[
x^2 – 5x + 4 = (x – 4)(x – 1)
\]

Final Answer:

\[
x^2 – 5x + 4 = (x – 4)(x – 1)
\]

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Algebra FAQ

Factor \(x^2-5x+4\).

\((x-1)(x-4)\).

Find the roots of \(x^2-5x+4=0\).

\(x=1\) or \(x=4\).

Solve \(x^2-5x+4=0\) by completing the square.

\(\left(x-\frac{5}{2}\right)^2=\frac{9}{4}\), so \(x=\frac{5}{2}\pm\frac{3}{2}\), giving \(1,4\).

What are the \(x\)-intercepts of \(y=x^2-5x+4\)?

Points \((1,0)\) and \((4,0)\).

Compute the vertex of \(y=x^2-5x+4\).

Vertex at \(x=\frac{-(-5)}{2\cdot 1}=\frac{5}{2}\). \(y=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)+4=-\frac{1}{4}\).

Determine if \(x^2-5x+4\) is always positive, always negative, or changes sign.

Discriminant \(25-16=9>0\) and leading coefficient is positive, so it changes sign at \(x=1\) and \(x=4\).

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