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# Polynomials and Radicals Practice Problems

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Updated on Oct 27, 2011

To review these concepts, go to Polynomials and Radicals Help.

## Polynomials and Radicals Practice Problems

Directions: Use scratch paper to solve the following problems. You can check your answers at the end of this section.

#### Practice

1. Simplify .
2. What is the product of 7x + 3 and 5x – 2?
3. Which of the following expressions can be factored using the rule a2b2 = (a + b)(ab)?
1. x2 – 27
2. x2 – 40
3. x2 – 48
4. x2 – 72
5. x2– 64
4. What are the factors of 56a5 + 21a?
5. If bx + 5c = 6adx, what does x equal in terms of a, b, c, and d?
6. Simplify .
7. What is a root of x(x – 1)(x + 1) = 27 – x?
8. What is one value that makes the fraction undefined?
9. What values make the fraction undefined?
10. What is one value that makes the fraction undefined?
11. 32x2 =
12. =
13. =
14. What is the value of (xy)(2xy)(3yx) if x = 2 and y = –2?

#### Solutions

1. To find the quotient:

2. Divide the coefficients and subtract the exponents.

3. To find the product, follow the FOIL method:

(7x + 3)(5x – 2)

F: 7x and 5x are the first pair of terms.

O: 7x and –2 are the outer pair of terms.

I: 3 and 5x are the inner pair of terms.

L: 3 and –2 are the last pair of terms.

Now multiply according to FOIL:

(7x)(5x) + (7x)(–2) + (3)(5x) + (3)(–2) = 35x2 – 14x + 15x – 6

Now combine like terms:

35x2 + x – 6

4. The rule a2b2 = (a + b)(ab) applies only to the difference between perfect squares, and 27, 40, 48, and 72 are not perfect squares. Choice e, 64, is a perfect square, so x2 – 64 can be factored as (x + 8)(x – 8).

5. To find the factors, determine a common factor for each term of 56a5 + 21a. Both coefficients (56 and 21) can be divided by 7 and both variables can be divided by a. Therefore, a common factor is 7a. Now, to find the second factor, divide the polynomial by the first factor:

6. Subtract exponents when dividing.

8a5 – 1 + 3a1 – 1

8a4 + 3a0

A base with an exponent of 0 = 1.

8a4 + 3(1)

8a4 + 3

Therefore, the factors of 56a5 + 21a are 7a(8a4 + 3).

7. Use factoring to isolate x:

First isolate the x terms on the same side.

8. bx + 5c + dx = 6adx + dx

bx + 5c + dx = 6a

bx + 5c + dx – 5c = 6a – 5c

bx + dx = 6a – 5c

Now factor out the common x term.

x(b + d) = 6a – 5c

Now divide to isolate x.

9. To simplify the expression, first determine the LCD of 8 and 10: The LCD is 40. Then convert each fraction into an equivalent fraction with 40 as the denominator:

Then simplify and reduce:

10. The root is 3. First, multiply the terms on the left side of the equation.

x(x –1) = x2x, (x2x)(x + 1) = x3 + x2x2x = x3x. Therefore, x3x = 27 – x. Add x to both sides of the equation: x3x + x = 27 – x +x, x3 = 27. The cube root of 27 is 3, so the root, or solution, of x(x – 1) (x + 1) = 27 – x is x = 3.

11. A fraction is undefined when its denominator is equal to 0. Set the denominator equal to 0 and solve for x: x3 + 125 = 0, x3 = –125, x = –5.
12. A fraction is undefined when its denominator is equal to 0. Factor the polynomial in the denominator and set each factor equal to 0 to find the values that make the fraction undefined: x3 + 3x2 – 4x = x(x + 4)(x – 1); x = 0; x + 4 = 0, x = –4; x – 1 = 0, x = 1. The fraction is undefined when x is equal to –4, 0, or 1.
13. A fraction is undefined when its denominator is equal to 0. Factor the polynomial in the denominator and set each factor equal to 0 to find the values that make the fraction undefined: 4x3 + 44x2 + 120x = 4x(x2 + 11x + 30) = 4x(x + 5)(x + 6); 4x = 0, x = 0; x + 5 = 0, x = –5; x + 6 = 0, x = –6. The fraction is undefined when x is equal to –6, –5, or 0.
14. Find the square root of the coefficient and the variable:√(32x2) = √32(x2) = x32. Next, factor √32 into two radicals, one of which is a perfect square: √32 = (√16)(√2) = 4√2. Therefore, √(32x2) = 4x2.
15. The cube root of 27y3 = 3y, because (3y)(3y)(3y) = 27y3. Factor the denominator into two radicals:√(27y2) = (√(9y2))(√3). The square root of 9y2 = 3y, because (3y)(3y) = 9y2. The expression is now equal to . Cancel the 3y terms from the numerator and denominator, leaving . Simplify the fraction by multiplying the numerator and denominator by
16. A term with a negative exponent can be rewritten as the reciprocal of the term with a positive exponent: . Square the numerator and denominator: . Therefore, .
17. First, multiply the first two terms. Multiply the coefficients of the terms and add the exponents. Because –y + y = 0, (xy)(2xy) = 2, and (xy)(2xy)(3yx) = 2(3yx) = 6yx. Substitute 2 for x and –2 for y: 6(–2)2 = 6(4) = 24.

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