Question

In the following exercises, factor each trinomial of the form

$$\begin{gathered} x^2+bxy+c{y}^2 \\ {m}^2-64mn-65n^2 \end{gathered}$$

Short answer

The binomial factors of the equations are$(m-65n)(m+n)$

Step-by-step solution

Step 1. Given information

The given equation is

$m^2-64mn-65n^2$

Step 2. Writing factors as two binomial with first term x and second term y.

The factors should be written as two binomials with the first term m and second term m.

$(mn)(mn)$

The product of the two numbers is 65, and their sum is 64. Because the third term sign is negative, the sum seems to be a difference.

$\left[\begin{array}{cc}\mathrm{Factors}& 65\\ & 1\\ \mathrm{Difference}& 64\end{array}\right]$

So form of binomial is$(m65n)(m1n)$.

Step 3. Finding factors  

The larger binomial second term is also negative because the second trinomial term is negative.

Because the third trinomial term is negative, the binomial signs are transposed.

The binomial factors are as follows:$(m-65n)(m+1n)$

Step 4. Checking the factors  

$$\begin{gathered} \mathrm{Check}:\mathrm{Multiply}\mathrm{the}\mathrm{factors}. \\ (m-65m)(m+n)=m^2+mn-65m^2-65mn\mathrm{True}\left[\mathrm{Multiply}\right] \\ =m^2-64mn-65m^2[\mathrm{Group}\mathrm{the}\mathrm{like}\mathrm{terms}] \end{gathered}$$