176.    x = –1

First, add 3 to each side of the equation.

9781118446614-eq180506.eps

Now raise each side of the equation to the third power.

4 – 4x = x 3 + 9x 2 + 27x + 27

Set the equation equal to 0 and solve for x.

0 = x 3 + 9x 2 + 31x + 23

You can use synthetic division to find a solution/factor; try x = –1.

9781118446614-eq180507.eps

x = –1 is a solution, so x + 1 is a factor of the polynomial. You can write x 3 + 9x 2 + 31x + 23 as (x + 1)(x 2 + 8x + 23), using the coefficients in the bottom row of the synthetic division.

The quadratic doesn’t factor, and, using the quadratic formula, you get imaginary

numbers. The imaginary/complex answers are 9781118446614-eq180508.eps.

So the only possible real answer is x = –1.

Check x = –1 using the original equation.

9781118446614-eq180509.eps

So the solution x = –1 works.

177.    x = 13

Square both sides of the equation.

9781118446614-eq180510.eps

Now add –5 to both sides.

9781118446614-eq180511.eps

Square both sides again.

x + 3 = 16

Solve for x.

x = 13

Check x = 13 in the ...

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