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Using Substitution Solving Polynomial Equations ?
Solve : (x3 + 4) 2 - 15 (x3 + 4) + 36 = 0.
You might be tempted to multiply everything out and factor. However, there's an easier way. Notice that this equation takes the form
A2 - 15A + 36 = 0,
where A = x3 + 4. You can easily solve this simpler quadratic equation! The left side factors to give
(A - 12)(A - 3) = 0
so A = 12 or A = 3.
Since A = x3 + 4, this is the same as saying
x3 + 4 = 12 or x3 + 4 = 3,
So x3 = 8 or x3 = -1,
which means
x = 2 or x = -1.
Normally, you can do this substitution of A for x3 + 4 in your head. You can simply write:
(x3 + 4) 2 - 15(x3 + 4) + 36 = 0
((x3 + 4) - 12)((x3 + 4) - 3) = 0
and so on.
Case 1: Suppose we have two terms 7ab and 3ab. When we multiply these two terms, we get 7ab x 3ab = (7 x 3) a 1 + 1 . b 1 + 1 ( Therefore, x m . x n = x m +
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