Unix and Equations
08.03 Factoring Trinomials x2 + bx + c
Part 1
Factor each trinomial below. Please show your work and check your answer. (1 point each)
x2 – 8x + 15
(x - 3) (x -5) x^2 - 5x - 3x +15 x^2 -8x + 15
a2 – a – 20
(a +4)(a-5) a^2 -5a +4a -20
a2 + 12ab + 27b2
(a +9b)(a +3b) a^2 + 3ab +9ab + 27b^2
2a2 + 30a + 100
(2a + 10)(a + 10)
2a^2 +20a +10a + 100
Part 2: (5 points)
It’s your turn to be a game show host! As you know, in the game of Math Time, the contestants are given an answer and they must come up with the question that corresponds to the given answer.
Your task for this portion of the assignment is to create two different “answers” (and the questions that accompany them) that the host could use for the final round of Math Time. The questions and answers you create must be unique. Check out the example and hint below, if needed.
x^2 - 100 is the product of these two binomials.
(x + 10) (x -10) x^2 -10x +10x -100
My Solution: c = current of river b = rate of boat d = s(t) will represent (distance = speed X time) Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations equal each other and solve for c: 10 + c = 10 - c 2c = 0 c = 0
The speed of the current was 0 mph Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 0 b = 10
The speed of the boat in still water must remain a consistent 10 mph or more in order for Wayne and his daughter to make it home in time or dinner.
My Solution: c = current of river b = rate of boat d = s(t) will represent (distance = speed X time) Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations
Part 1
Factor each trinomial below. Please show your work and check your answer. (1 point each)
x2 – 8x + 15
(x - 3) (x -5) x^2 - 5x - 3x +15 x^2 -8x + 15
a2 – a – 20
(a +4)(a-5) a^2 -5a +4a -20
a2 + 12ab + 27b2
(a +9b)(a +3b) a^2 + 3ab +9ab + 27b^2
2a2 + 30a + 100
(2a + 10)(a + 10)
2a^2 +20a +10a + 100
Part 2: (5 points)
It’s your turn to be a game show host! As you know, in the game of Math Time, the contestants are given an answer and they must come up with the question that corresponds to the given answer.
Your task for this portion of the assignment is to create two different “answers” (and the questions that accompany them) that the host could use for the final round of Math Time. The questions and answers you create must be unique. Check out the example and hint below, if needed.
x^2 - 100 is the product of these two binomials.
(x + 10) (x -10) x^2 -10x +10x -100
My Solution: c = current of river b = rate of boat d = s(t) will represent (distance = speed X time) Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations equal each other and solve for c: 10 + c = 10 - c 2c = 0 c = 0
The speed of the current was 0 mph Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 0 b = 10
The speed of the boat in still water must remain a consistent 10 mph or more in order for Wayne and his daughter to make it home in time or dinner.
My Solution: c = current of river b = rate of boat d = s(t) will represent (distance = speed X time) Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations