Question

1. $x^2 + 7xy + 6y^2$

Explanation:

Step1: Analyze the quadratic trinomial

We have the expression \(x^{2}+7xy + 6y^{2}\). We need to factor it. For a quadratic of the form \(ax^{2}+bx + c\) (here, considering \(x\) as the variable and \(y\) as a constant coefficient for the terms involving \(x\)), we need to find two numbers that multiply to \(a\times c\) (here \(a = 1\), \(c=6y^{2}\), so the product is \(6y^{2}\)) and add up to \(b\) (here \(b = 7y\)).
The two numbers that multiply to \(6y^{2}\) and add up to \(7y\) are \(y\) and \(6y\) since \(y\times6y=6y^{2}\) and \(y + 6y=7y\).

Step2: Rewrite the middle term

Rewrite the middle term \(7xy\) as \(xy+6xy\). So the expression becomes:
\(x^{2}+xy + 6xy+6y^{2}\)

Step3: Group the terms

Group the first two terms and the last two terms:
\((x^{2}+xy)+(6xy + 6y^{2})\)

Step4: Factor out the common factors from each group

From the first group \(x^{2}+xy\), we can factor out \(x\) to get \(x(x + y)\).
From the second group \(6xy+6y^{2}\), we can factor out \(6y\) to get \(6y(x + y)\).
So now the expression is:
\(x(x + y)+6y(x + y)\)

Step5: Factor out the common binomial factor

Now we can factor out \((x + y)\) from both terms:
\((x + y)(x + 6y)\)

Answer:

\((x + y)(x + 6y)\)