Q. \(x^2+6x+8\).
Answer
Factor the quadratic to find its zeros.
\[
x^2+6x+8=(x+2)(x+4)
\]
So \(x+2=0\) or \(x+4=0\), giving \(x=-2\) or \(x=-4\).
\[
\boxed{x=-2,\,-4}
\]
Detailed Explanation
We want to simplify the expression \(x^2+6x+8\) and understand its structure.
Step 1: Identify how to factor the quadratic.
The expression \(x^2+6x+8\) is a quadratic polynomial of the form \(x^2+bx+c\) with \(b=6\) and \(c=8\).
Step 2: Find two numbers that multiply to \(8\) and add to \(6\).
We look for numbers \(m\) and \(n\) such that:
- \(mn=8\)
- \(m+n=6\)
The pair \(2\) and \(4\) works because:
- \(2\cdot 4=8\)
- \(2+4=6\)
Step 3: Factor the quadratic using those numbers.
Now we rewrite the quadratic as a product of two binomials:
\[
x^2+6x+8=(x+2)(x+4)
\]
Final Answer:
\[
x^2+6x+8=(x+2)(x+4)
\]
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Algebra FAQ
-
Factor \(x^2+6x+8\).
\(x^2+6x+8=(x+2)(x+4)\).
-
Find the roots of \(x^2+6x+8=0\).
\((x+2)(x+4)=0\) so \(x=-2\) or \(x=-4\).
-
Complete the square for \(x^2+6x+8\).
\(x^2+6x+8=(x+3)^2-1\).
-
Determine the vertex and minimum value of \(y=x^2+6x+8\).
Vertex at \(x=-3\). \(y_{\min}=( -3+3)^2-1=-1\).
-
Is \(x^2+6x+8\) positive, negative, or zero for real \(x\)?
Since roots are \(-4\) and \(-2\), it is \(=0\) at those points, negative on \((-4,-2)\), and positive outside that interval.
-
Compute \(x^2+6x+8\) at \(x=-2\).
\((-2)^2+6(-2)+8=4-12+8=0\).
-
Compute the discriminant of \(x^2+6x+8\).
\(a=1,b=6,c=8\). \(\Delta=b^2-4ac=36-32=4\).
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