Q. \(x^2-3x-4\)

Q: Solve x^2-3x-4=0 by factoring.

\;x^2-3x-4=(x-4)(x+1)\Rightarrow x=4 or x=-1.

Q: Solve x^2-3x-4=0 using the quadratic formula.

\;a=1,b=-3,c=-4. x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-4)}}{2(1)}=\frac{3\pm5}{2}\Rightarrow x=4,-1.

Q: Find the roots and check them in the original polynomial.

Test x=4: 16-12-4=0. Test x=-1: 1+3-4=0. Both work.

Q: What is the vertex of y=x^2-3x-4?

Vertex at x=\frac{-b}{2a}=\frac{3}{2}. y=\left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)-4=\frac{9}{4}-\frac{9}{2}-4=-\frac{25}{4}.

Q: Determine the y-intercept and x-intercepts.

Y-intercept: x=0\Rightarrow y=-4. X-intercepts: solve roots x=4 and x=-1.

Q: Find the discriminant and interpret it.

\Delta=b^2-4ac=9+16=25. Two distinct real roots because \Delta>0.

Answer

We factor the quadratic by finding two numbers that multiply to \(-4\) and add to \(-3\). Those numbers are \(-4\) and \(1\).

\[
x^2 – 3x – 4 = (x-4)(x+1)
\]

So the factored form is \((x-4)(x+1)\).

Detailed Explanation

We want to work with the expression \(x^2 – 3x – 4\). A common first step is to factor it (when possible), because factoring helps with simplifying, solving equations, and understanding the structure of the polynomial.

Step 1: Identify what we are factoring

The expression is a quadratic polynomial:

\[
x^2 – 3x – 4
\]

To factor a quadratic of the form \(ax^2 + bx + c\), we look for two numbers that multiply to \(ac\) and add to \(b\).

Here, \(a = 1\), \(b = -3\), and \(c = -4\).

Step 2: Find numbers that multiply to \(a \cdot c\)

Compute \(a \cdot c\):

\[
1 \cdot (-4) = -4
\]

So we need two numbers whose product is \(-4\).

Step 3: Find numbers that add to \(b\)

We also need those same two numbers to add to \(b = -3\).

Check pairs that multiply to \(-4\):

\(1\) and \(-4\) give product \(-4\), but sum is \(-3\). That matches our target.

Step 4: Write the factorization

Since the numbers are \(1\) and \(-4\), the factored form is:

\[
x^2 – 3x – 4 = (x + 1)(x – 4)
\]

Final Answer

The expression \(x^2 – 3x – 4\) factors as:

\[
x^2 – 3x – 4 = (x + 1)(x – 4)
\]

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Algebra FAQ

Solve \(x^2-3x-4=0\) by factoring.

\(\;x^2-3x-4=(x-4)(x+1)\Rightarrow x=4\) or \(x=-1\).

Solve \(x^2-3x-4=0\) using the quadratic formula.

\(\;a=1,b=-3,c=-4\). \(x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-4)}}{2(1)}=\frac{3\pm5}{2}\Rightarrow x=4,-1\).

Find the roots and check them in the original polynomial.

Test \(x=4\): \(16-12-4=0\). Test \(x=-1\): \(1+3-4=0\). Both work.

What is the vertex of \(y=x^2-3x-4\)?

Vertex at \(x=\frac{-b}{2a}=\frac{3}{2}\). \(y=\left(\frac{3}{2}\right)^2-3\left(\frac{3}{2}\right)-4=\frac{9}{4}-\frac{9}{2}-4=-\frac{25}{4}\).

Determine the y-intercept and x-intercepts.

Y-intercept: \(x=0\Rightarrow y=-4\). X-intercepts: solve roots \(x=4\) and \(x=-1\).

For what \(x\) is \(x^2-3x-4\) positive?

Since leading coefficient is positive, polynomial is positive outside the roots: \(x<-1\) or \(x>4\).

Find the discriminant and interpret it.

\(\Delta=b^2-4ac=9+16=25\). Two distinct real roots because \(\Delta>0\).

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