Now here’s the trick! If we spend enough time playing around with this, we’ll find that it can be factored into(3x+ 2)(x^2 + 1) = 0When we set the first binomial equal to 0, we get3 x+ 2 = 0which resolves to x = −2/3. When we set the second binomial equal to 0, we getx^2 + 1 = 0Subtracting 1 from both sides gives usx^2 =− 1This resolves to x = j or x = −j. We have found three roots to the cubic we got by mixing:x=−2/3 or x=j or x=−jWe can substitute these roots into either of the original cubics. Let’s use the second one. For x = −2/3,
we havey= 2 x^3 +x^2 + 2 x+ 5
= 2 × (−2/3)^3 + (−2/3)^2 + 2 × (−2/3)+ 5
= 2 × (−8/27)+ 4/9 + (−4/3)+ 5
=−16/27+ 12/27 + (−36/27)+ 135/27
= 95/27Our first solution is (x,y) = (−2/3, 95/27). For x = j, we havey= 2 x^3 +x^2 + 2 x+ 5
= 2 j^3 +j^2 + 2 j+ 5
= 2(−j)+ (−1)+j 2 + 5
=−j 2 + (−1)+j 2 + 5
=−j 2 +j 2 + (−1)+ 5
= 0 + 4
= 4Our second solution is (x,y) = (j, 4). For x = −j, we havey= 2 x^3 +x^2 + 2 x+ 5
= 2(−j)^3 + (−j)^2 + 2(−j )+ 5
= 2 j+ (−1)+ (−j2)+ 5460 More Two-by-Two Systems