Can someone help me with these I am so lost and caonfused.?

Question

1.Subtract 4x^2 – 5x from the sum of 9x^2 – 2 and 6x – 1.

2. The profit on a watch is given by P = x^2 – 13x – 80, where x is the number of watches sold per day. How many watches were sold on a day when there was a $50 loss?

3. Use the ac test to determine if the following trinomial can be factored.
x^2 + 2x + 3
4. State which method should be applied as the first step for factoring the polynomial.
2a^2 + 9a + 10

Answer 1

1. (9x^2 – 2) + (6x – 1) - (4x^2 – 5x) =
9x^2 - 4x^2 + 6x - (-5x) - 2 - 1 =
5x^2 + 11x - 3

2. P = x^2 – 13x – 80
P = -50 on this day.

Thus:

-50 = x^2 - 13x - 80
x^2 - 13x - 30 = 0
(x - 15)(x + 2) = 0

x = 15, or x = -2, are both solutions. But you can't sell -2 watches (unless maybe two previous day's buyers returned watches?).

3. x^2 + 2x + 3
By the "ac test" I presume they want the discriminant (b^2 - 4ac):

b^2 - 4ac =
2^2 - 4*3 =
4 - 12 =
-8

The equation has no real solutions, since the discriminant is negative.

4. 2a^2 + 9a + 10
That one looks pretty easy to me.

You need two factors of 10, where twice one factor plus the other equals nine. For 5,2 that works (5 + 2*2 = 9). This leads to this factoring:

(2a + 5)(a + 2)

Answer 2

1.Subtract 4x^2 – 5x from the sum of 9x^2 – 2 and 6x – 1.

9x^2 – 2 + 6x – 1
= 9x^2 + 6x - 3 - (4x^2 - 5x)
= 9x^2 + 6x - 3 - 4x^2 + 5x
= 5x^2 + 11x - 3


2. P = x^2 – 13x – 80
-50 = x^2 – 13x – 80
0 = x^2 – 13x – 80 + 50
0 = x^2 -13x - 30
0 = (x -15)(x + 2) -----------------> Set both of these to 0.
0 = (x - 15) & 0 = (x + 2) -----------------> Solve for x.

x = 15 watches.... Is the number that were sold.
Because x = -2 also, but you can't sell a negative number of items.

3. Use the ac test to determine if the following trinomial can be factored.

x^2 + 2x + 3 ---------------> given equation
(ax^2 + bx + c)

a = 1
b = 2
c = 3

a* c = 1 * 3 = 3
a + c = b
1 + 3 = 4 not 2

Not factorable.


4. State which method should be applied as the first step for factoring the polynomial.
2a^2 + 9a + 10

Brackets and putting a 2a in the first spot.
(2a )(a )

Then I would find what is going in the last places...
(2a + 5)(a + 2) and see if they work...

Answer 3

1) 9x^2-2 + 6x-1 = 9x^2+6x-3
9x^2+6x-3 -(4x^2-5x) = 5x^2+11x-3

2)-50=x^2-13x-80
x^2-13x-30=0
x=(13+/-sqrt(169+120))/2 = 15 and -2
since -2 watches in not a valid answer, the answer is 15 watches

3) This can't be factored, since there are no set of integers that have a product of 3 and a sum of 2

4) AC method

Answer 4

1. (9x^2-2)+(6x-1)-(4x^2-5x) = 5x^2+11x-3

2. -50= x^2-13x-80 when x is 15, you get -50.

3. Cannot be factored. There are no numbers that mult to make 3 and add up to 2

4. take A times C.

Answer 5

1)The sum is 9x^2 +6x-3
Then the answer is simply
(9x^2 +6x-3) - (4x^2 - 5x)
=5x^2 +11x +2
Make sure you distribute the proper signs.

2)Well, a $50 loss means P= -50, so
-50 = x^2 -13x - 80.
0=x^2 - 13x - 30
Use quadratic formula to solve,
x= (1/2)*( 13 +/- sqrt(169 + 120))
x = (1/2)*(13 +/- sqrt(289))
x = (1/2)*(13 +/- 17)
x = 15 or x = -2
The only reasonable answer is x= 15 because you won't sell a negative number of watches.

Answer 6

1. Your base equation will look like the following:
((9x^2-2)(6x-1))-(4x^2-5x)
(54x^3-9x^2-12x+2)-(4x^2-5x)
54x^3-9x^2-12x+2-4x^2+5x
54x^3-13x^2+7x+2
2. P=x^2-13x-80 with P=-50
x^2-13x-80=-50
solve for 0 to get
x^2-13x-30=0
(x-15)(x+2)=0
x=15 or -2