Q. \(x^2+7x+6\)

Q: What are the zeros of x^2+7x+6?

Factor x^2+7x+6=(x+1)(x+6). So the zeros are x=-1 and x=-6.

Q: How do you factor x^2+7x+6?

Find numbers multiplying to 6 and adding to 7: 1 and 6. Thus x^2+7x+6=(x+1)(x+6).

Q: Solve x^2+7x+6=0.

Using the factorization (x+1)(x+6)=0, get x=-1 or x=-6.

Q: What is the vertex of y=x^2+7x+6?

For ax^2+bx+c, x=-\frac{b}{2a}=-\frac{7}{2}. Then y=\left(-\frac{7}{2}\right)^2+7\left(-\frac{7}{2}\right)+6=-\frac{1}{4}.

Q: What are the y-intercept and x-intercepts?

y-intercept: x=0\Rightarrow y=6. x-intercepts: set y=0\Rightarrow x=-1,-6.

Q: What is the minimum value of the quadratic?

Since a=1>0, the minimum is at the vertex: -\frac{1}{4}.

Answer

We factor the quadratic:

\[
x^2+7x+6=(x+1)(x+6)
\]

So the solutions come from each factor being zero:

\[
x+1=0 \Rightarrow x=-1,\quad x+6=0 \Rightarrow x=-6
\]

Final result: \(x=-1\) or \(x=-6\).

Detailed Explanation

We want to simplify or factor the expression \(x^2+7x+6\). A common goal with a quadratic like this is to factor it into \((x+a)(x+b)\).

Step 1: Identify the quadratic form

A quadratic polynomial has the form

\[x^2+7x+6\]

Here, the coefficient of \(x^2\) is \(1\), the coefficient of \(x\) is \(7\), and the constant term is \(6\).

Step 2: Factor as \((x+a)(x+b)\)

To factor \(x^2+7x+6\), we look for numbers \(a\) and \(b\) such that

\[(x+a)(x+b)=x^2+(a+b)x+ab\]

So we need

\[a+b=7\]

and

\[ab=6\]

Step 3: Find two numbers with sum \(7\) and product \(6\)

List factor pairs of \(6\):

\[1\cdot6=6,\quad 2\cdot3=6\]

Check sums:

\[1+6=7,\quad 2+3=5\]

Only \(1\) and \(6\) add to \(7\).

Step 4: Write the factored form

Thus, \(a=1\) and \(b=6\). Therefore,

\[x^2+7x+6=(x+1)(x+6)\]

Final Answer

\[x^2+7x+6=(x+1)(x+6)\]

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Algebra FAQ

What are the zeros of \(x^2+7x+6\)?

Factor \(x^2+7x+6=(x+1)(x+6)\). So the zeros are \(x=-1\) and \(x=-6\).

How do you factor \(x^2+7x+6\)?

Find numbers multiplying to \(6\) and adding to \(7\): \(1\) and \(6\). Thus \(x^2+7x+6=(x+1)(x+6)\).

Solve \(x^2+7x+6=0\).

Using the factorization \((x+1)(x+6)=0\), get \(x=-1\) or \(x=-6\).

What is the vertex of \(y=x^2+7x+6\)?

For \(ax^2+bx+c\), \(x=-\frac{b}{2a}=-\frac{7}{2}\). Then \(y=\left(-\frac{7}{2}\right)^2+7\left(-\frac{7}{2}\right)+6=-\frac{1}{4}\).

What are the \(y\)-intercept and \(x\)-intercepts?

\(y\)-intercept: \(x=0\Rightarrow y=6\). \(x\)-intercepts: set \(y=0\Rightarrow x=-1,-6\).

What is the minimum value of the quadratic?

Since \(a=1>0\), the minimum is at the vertex: \(-\frac{1}{4}\).

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