Algebra Know-It-ALL

We can add y to each side and transpose the equation left-to-right, getting the function

y= 3 x+ 1

Next, we mix

When we mix the independent-variable parts of the above functions, we obtain one equation
in one variable:

x^3 + 6 x^2 + 14 x+ 7 = 3 x+ 1

If we subtract the quantity (3x+ 1) from both sides, we get

x^3 + 6 x^2 + 11 x+ 6 = 0

This is a straightforward cubic equation in polynomial standard form. The roots aren’t obvi-
ous from casual inspection, but we can use the techniques from Chap. 25 to solve it.

Next, we solve

Now that we have derived a cubic equation in one variable, our mission is to find its roots. We
can use synthetic division several times to obtain factors. Ultimately, we find that the cubic
factors into

(x+ 1)(x+ 2)(x+ 3) = 0

The roots can be found by solving the three equations we get when we set each binomial equal
to 0. Those roots are x=−1,x=−2, and x=−3. The y-values can be found by plugging these
roots into either of the original functions. Let’s use the linear one; it’s the less messy of the
two! For x=−1, we have

y= 3 x+ 1

= 3 × (−1)+ 1

=− 3 + 1

=− 2

Now we know that our first solution is (x,y)= (−1,−2). When x=−2, we have

y= 3 x+ 1

= 3 × (−2)+ 1

=− 6 + 1

=− 5

Enter the Cubic 457