Skip to Main Content
1001 Algebra II Practice Problems For Dummies
book

1001 Algebra II Practice Problems For Dummies

by Mary Jane Sterling
June 2013
Beginner content levelBeginner
504 pages
9h 8m
English
For Dummies
Content preview from 1001 Algebra II Practice Problems For Dummies

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 ...

Become an O’Reilly member and get unlimited access to this title plus top books and audiobooks from O’Reilly and nearly 200 top publishers, thousands of courses curated by job role, 150+ live events each month,
and much more.
Start your free trial

You might also like

Basic Technical Mathematics, 11th Edition

Basic Technical Mathematics, 11th Edition

Allyn J. Washington, Richard Evans

Publisher Resources

ISBN: 9781118446614Publisher Support