# Solve Quadratic-Like Trinomials

If you ever have to solve a quadratic-like trinomial, you’re in luck; this type of equation is a perfect candidate for factoring and then for the application of the multiplication property of zero.

A quadratic-like trinomial is a trinomial of the form *ax*^{2}* ^{n}* +

*bx*

*+*

^{n}*c*= 0. The power on one variable term is twice that of the other variable term, and a constant term completes the picture.

The *multiplication property of zero* states that if the product of

then at least one of the factors has to represent the number 0.

Now, try an example: Solve the trinomial *z*^{6} – 26*z*^{3} – 27 = 0.

You can think of this equation as being like the quadratic

*x*^{2} – 26*x* – 27,

which factors into

(*x* – 27)(*x* + 1).

If you replace the *x*‘s in the factorization with *z*^{3}, you have the factorization for the equation with the *z*‘s.

*z*^{6} – 26*z*^{3} – 27 = (*z*^{3} – 27)(*z*^{3} + 1) = 0

Then you then set each factor equal to zero. When *z*^{3} – 27 = 0, *z*^{3} = 27, and *z* = 3. And when *z*^{3} + 1 = 0, *z*^{3} = –1, and *z* = –1.

You can just take the cube roots of each side of the equations you form, because when you take that odd root, you know you can find only one real solution.

Here’s another example. When solving the quadratic-like trinomial

*y*^{4} – 17*y*^{2} + 16 = 0,

you can factor the left side and then factor the factors again:

*y*^{4} – 17*y*^{2} + 16 = (*y*^{2} – 16)(*y*^{2} – 1) = (*y* – 4)(*y* + 4)(*y* – 1)(*y* + 1) = 0.

Setting the individual factors equal to zero, you get *y* = 4, *y* = –4, *y* = 1, *y* = –1.