Q. \(x^2+5x+4\)

Answer

\(x^2+5x+4\) factors as \((x+1)(x+4)\) since \(1\cdot 4=4\) and \(1+4=5\).

Final result: \((x+1)(x+4)\)

Detailed Explanation

We want to work with the polynomial \(x^2 + 5x + 4\). A common and useful step is to factor it, because factoring reveals the roots and helps with other algebra tasks.

Step 1: Identify the form to factor

The expression \(x^2 + 5x + 4\) is a quadratic of the form \(ax^2 + bx + c\), where:

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

Step 2: Factor using the “find two numbers” method

We look for two numbers \(m\) and \(n\) such that:

  • \(m + n = 5\)
  • \(mn = 4\)

Step 3: Choose the correct numbers

List the factor pairs of \(4\):

  • \(1 \cdot 4 = 4\)
  • \((-1) \cdot (-4) = 4\)

Now check sums:

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

The pair that gives the sum \(5\) is \(1\) and \(4\).

Step 4: Write the factored form

Substitute these values into the factor structure:

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

Final Answer

The expression factors as:

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

See full solution

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Algebra FAQ

Factorize \(x^2+5x+4\).

Find numbers multiplying to \(4\) and adding to \(5\): \(1\) and \(4\). So \(x^2+5x+4=(x+1)(x+4)\).

Solve \(x^2+5x+4=0\).

Use factors: \((x+1)(x+4)=0\). Thus \(x=-1\) or \(x=-4\).

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

Set \(y=0\): \(x=-1,-4\). Intercepts: \((-1,0)\) and \((-4,0)\).

Complete the square for \(x^2+5x+4\).

\(x^2+5x+4=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}+4=\left(x+\frac{5}{2}\right)^2-\frac{9}{4}\).

Determine the vertex and minimum value of \(y=x^2+5x+4\).

Vertex at \(x=-\frac{b}{2a}=-\frac{5}{2}\). Minimum \(y=-\frac{9}{4}\). Vertex: \(\left(-\frac{5}{2},-\frac{9}{4}\right)\).

Compute the discriminant of \(x^2+5x+4=0\).

\(a=1,b=5,c=4\). Discriminant \(D=b^2-4ac=25-16=9\), so two real roots.
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