Factorng word problem- help please!?

Question

1. A person purchased 5k + 2 items for a total cost of 35k^2 + 29k +6. Find the average cost per item of this purchase.

2. Another person bought a certain number of an item at a cost of 4k + 3 each/ Find the number bought if the total cost was 8k^2 +2k - 3.

Thanks to anyone who can help with the factoring.. it confuses me so much.

Answer 1

1.(35k^2+29k+6)/(5k+2)
=(5k+2)(7k+3)/(5k+2)[factoring dividend]
=7k+3 [elliminating 5k+2]
The average cost is 7k+3
2.(8k^2+2k-3)/(4k+3)
=(4k+3)(2k-1)/(4k+3)[factoring dividend]
=2k-1[elliminating 4k+3]
Therefore he purchased 2k-1 items

Answer 2

1. A person purchased 5k + 2 items for a total cost of 35k^2 + 29k +6. Find the average cost per item of this purchase.

Ans: Avg. Cost of each item = (35k^2 + 29k + 6) / (5k + 2)

= (5k + 2) (7k + 3) / (5k + 2) = 7k + 3

2. Another person bought a certain number of an item at a cost of 4k + 3 each/ Find the number bought if the total cost was 8k^2 +2k - 3.

Ans: Number of items bought = Total cost / cost of each

8k^2 + 2k - 3 / 4k + 3

(4k + 3) (2k - 1) / (4k + 3) = 2k - 1

The trick is to factorise the quadratic equations and it is good to assume that the denominator is designed to be one of the factors.

Answer 3

1. the other factor must have 7k (so that 5k times 7k = 35k^2 that you need) and also must have a 3 (so that 2 time 3 = 6 that you need). Thus the factor is (7k + 3). Check
(5k + 2)(7k + 3) = 35k^2 + 15k + 14k + 6 which is
35k^2 + 29k + 6

2. Same process gets you the factor of (2k - 1)

Answer 4

Factoring polynomials like these are hard to see, but there is a better way: Long division!
How many 5k's will go into 35k^2?
That would be 7k
Thus multiply 7k by 5k+2 to get 35k^2 + 14k.
Place this underneath 35k^2 + 29k + 6 and subtract.
The result is: 15k + 6
How many 5k's go into 15k?
That would be 3
Thus multiply 3 by 5k+2 to get 15k + 6.
Place this underneath the 15k + 6 we had from the previous step and subtract.
The remainder is zero!
Thus (35k^2 + 29k + 6) / (5k + 2) = 7k + 3

When you do long division on polynomials and get a zero remainder, you have taken a factor out of it and are well on your way to a complete factorization. Have fun!

Answer 5

1. (35k^2 + 29k + 6) / (5k+2) =

(5k+2)(7k + 3) / (5k+2) = 7k+3

2. (8k^2 + 2k -3) / (4k+3) =

(4k+3)(2k-1) / (4k+3) = 2k-1

Answer 6

35k^2 + 29k + 6 / 5k + 2

. . . . . . . .. . . .. .7 + 3
5 + 2 // 35 + 29 + 6
. . . . . . .35 + 14
. . . . . .. _______
. . . . . .. . . . . 15 + 6
. . . . . .. . . . . 15 + 6

Answer is 7x + 3

. . . . . . . . . .2 - 1
4 + 3 // 8 + 2 - 3
. . . . . . .8 + 6
.. . . . . ._____
. . . . . . . . . -4 -3
. . . . . . . . . -4 -3

2x - 1

Answer 7

1. 7k+2+2/(5k+2)
2.2k+1-6/(4k+3)