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Question

Directions work each problem

 

1. (2y³+2y²-y+16) + (5y³+3y-3)

 

2. 5(x+y) -4(3x-2y+1)

 

3. 4^-3

 

Factor 

4.  9x²-49

 

5.  x²+14x+33

Answer 1

1.  remove parentheses & collect like terms 9because there's nothing to distribute into the parenthesized quantities.

2.  distribute & collect like terms

3. convert from negative exponent to reciprocal: = 1/(4³) = 1/64

4. = (3x)² - 7² & now use DIFFERENCE OF SQUARES to factor; that principle must be memorized & always be available

5. = (x+11)(x+3) ... you had to split the 33 into two factors whose sum is 14, the coefficient of the linear term.

 

Study hard & learn, so that soon you can do your homework on your own.

Answer 2

1.  (2y³+2y²-y+16) + (5y³+3y-3)

 

Since there are no numbers in front of the parenthesis,

there is nothing to distribute through.  This means this

is a straight addition problem.  All you do is combine like terms.

 

(2y³+2y²-y+16) + (5y³+3y-3) = 2y³+2y²-y+16+5y³+3y-3 =

 

Let's put the terms with same variable next to each other in

descending order from highest exponent to lowest.

 

2y³+5y³+2y²-y+3y+16-3

 

= 7y³ + 2y² + 2y + 13

 

2. 5(x+y) -4(3x-2y+1)

 

Here we have a number or expression in front of the parenthesis

so we need to distribute those through and then combine like

terms as we did in problem 1.  To distribute through the

parenthesis, each term inside the parenthesis is multiplied by

the number or term in front of the parenthesis.

 

5(x+y)- 4(3x-2y+1)

= 5(x) + 5(y) -4(3x)-4(-2y)-4(1)

= 5x+5y-12x+8y-4

Now just combine the  terms that have the same variables.

 

= 5x-12x+5y+8y-4

 

= -7x + 13y - 4

 

3.   4^-3

 

With exponents note that a^-n = 1/aⁿ  and 1/a^-n = aⁿ

 

Example:  x^-3 = 1/x³     and   1/x^-3 = x³

 

4^-3 = 1/4³ = 1/[4(4)(4)] = 1/64

 

4.  Factor 9x² - 49

 

This is the difference of two square terms.

The basic formula is  (a²-b²) = (a-b)(a+b)

 

9x² = (3x)²

49 = 7²

Thus for the formula  a=3x ... b = 7

 

9x²-49 = (3x-7)(3x+7)

 

5.  Factor:  x²+14x+33

 

Since there is no coefficient on the square term, to factor

this we look for factors of 33 that will add to the coefficient

of the x term.   We need factors of 33 that add to 14.

 

(11)(3) = 33  and  11 + 3 = 14

So the numbers we need are 11 and 3

 

x² + 14x + 33 = (x+11)(x+3)

 

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Note::  What if this had been a problem with a negative?

Example: x² + 8x - 33

 

Here we need factors of -33 that add to 8

11(-3) = -33   and  11-3 = 8 so...

 

x² + 8x - 33 = (x+11)(x-3)