2.6 Other types of equations  (Page 2/10)

First, set the equation equal to zero. Then factor out what is common to both terms, the GCF.

Notice that we have the difference of squares in the factor $x^2-16,$ which we will continue to factor and obtain two solutions. The first term, $5x^2,$ generates, technically, two solutions as the exponent is 2, but they are the same solution.

The solutions are $x=0\text{ (double solution),}$ $x=4,$ and $x=\mathrm{-4.}$

This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.

The grouping process ends here, as we can factor $x^2-9$ using the difference of squares formula.

The solutions are $x=3,$ $x=\mathrm{-3},$ and $x=\mathrm{-1.}$ Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x- intercepts, on the graph in [link] .