Divide.
(20x2-12x+8) ÷(2x+8)
(1 point)
11 answers
10x - 6
Sorry, I can not do it.
No problem! Here is the solution:
To divide (20x^2 - 12x + 8) by (2x + 8), we need to use polynomial long division. Here is the process:
10x - 6
_______________________
2x + 8 | 20x^2 - 12x + 8
- (20x^2 + 80)
________________
-92x + 8
-(-92x - 368)
______________
376
Therefore, (20x^2 - 12x + 8) ÷ (2x + 8) = 10x - 6 + 376 / (2x + 8).
To divide (20x^2 - 12x + 8) by (2x + 8), we need to use polynomial long division. Here is the process:
10x - 6
_______________________
2x + 8 | 20x^2 - 12x + 8
- (20x^2 + 80)
________________
-92x + 8
-(-92x - 368)
______________
376
Therefore, (20x^2 - 12x + 8) ÷ (2x + 8) = 10x - 6 + 376 / (2x + 8).
Hummm, I suspect the student has a typo.
Ah, I see it now. Thank you for pointing that out. Let's try it again assuming it was a typo in the expression:
(20x^2 - 12x + 8) ÷ (2x + 8)
Factor out a 2 from the first two terms in the numerator:
= 2(10x^2 - 6x + 4) ÷ (2x + 8)
Now we simplify the expression inside the parentheses:
= 2(5x - 2)(2x + 4) ÷ (2x + 8)
Factor out the common factors:
= 2(5x - 2)(2x + 4) ÷ 2(x + 4)
= (5x - 2)(2x + 4)
Therefore, (20x^2 - 12x + 8) ÷ (2x + 8) simplifies to (5x - 2)(2x + 4).
(20x^2 - 12x + 8) ÷ (2x + 8)
Factor out a 2 from the first two terms in the numerator:
= 2(10x^2 - 6x + 4) ÷ (2x + 8)
Now we simplify the expression inside the parentheses:
= 2(5x - 2)(2x + 4) ÷ (2x + 8)
Factor out the common factors:
= 2(5x - 2)(2x + 4) ÷ 2(x + 4)
= (5x - 2)(2x + 4)
Therefore, (20x^2 - 12x + 8) ÷ (2x + 8) simplifies to (5x - 2)(2x + 4).
sorry
(10x^2 - 6x + 4)
is not
(5x - 2)(2x + 4)
(10x^2 - 6x + 4)
is not
(5x - 2)(2x + 4)
Let's try the division again with the correct coefficients:
(20x^2 - 12x + 8) ÷ (2x + 8)
First, divide 20x^2 by 2x to get 10x. Multiply 2x by 10x to get 20x^2. Multiply 10x by 8 to get 80x. Subtract this from -12x to get -92x. Write this below the line:
10x -6
______________
2x + 8 | 20x^2 -12x + 8
- (20x^2 +80x)
______________
-92x + 8
Therefore, (20x^2 - 12x + 8) ÷ (2x + 8) equals 10x - 6 with a remainder of -92x + 8.
(20x^2 - 12x + 8) ÷ (2x + 8)
First, divide 20x^2 by 2x to get 10x. Multiply 2x by 10x to get 20x^2. Multiply 10x by 8 to get 80x. Subtract this from -12x to get -92x. Write this below the line:
10x -6
______________
2x + 8 | 20x^2 -12x + 8
- (20x^2 +80x)
______________
-92x + 8
Therefore, (20x^2 - 12x + 8) ÷ (2x + 8) equals 10x - 6 with a remainder of -92x + 8.
So I think there is a typo in the problem statement.
Ah, I see. Thank you for pointing that out. Let's correct the expression and solve it again:
Given expression: (20x^2 - 12x + 8) ÷ (2x + 8)
First, let's factor out a 4 from all parts of the polynomial to simplify the division:
(4(5x^2 - 3x + 2)) ÷ (2(x + 4))
Now we have:
(4(5x^2 - 3x + 2)) ÷ (2(x + 4))
= 4(5x^2 - 3x + 2) ÷ 2(x + 4)
= 2(5x^2 - 3x + 2) ÷ (x + 4)
Now, we can divide the polynomial 5x^2 - 3x + 2 by x + 4 using polynomial division:
5
___________
x + 4 | 5x^2 - 3x + 2
- (5x^2 + 20)
_______________
-23x + 2
Therefore, the division simplifies to 5 with a remainder of -23x + 2, and the expression becomes:
2(5) - (23x + 2) / (x + 4)
= 10 - (23x + 2) / (x + 4)
So, the simplified form of (20x^2 - 12x + 8) ÷ (2x + 8) is 10 - (23x + 2) / (x + 4)
Given expression: (20x^2 - 12x + 8) ÷ (2x + 8)
First, let's factor out a 4 from all parts of the polynomial to simplify the division:
(4(5x^2 - 3x + 2)) ÷ (2(x + 4))
Now we have:
(4(5x^2 - 3x + 2)) ÷ (2(x + 4))
= 4(5x^2 - 3x + 2) ÷ 2(x + 4)
= 2(5x^2 - 3x + 2) ÷ (x + 4)
Now, we can divide the polynomial 5x^2 - 3x + 2 by x + 4 using polynomial division:
5
___________
x + 4 | 5x^2 - 3x + 2
- (5x^2 + 20)
_______________
-23x + 2
Therefore, the division simplifies to 5 with a remainder of -23x + 2, and the expression becomes:
2(5) - (23x + 2) / (x + 4)
= 10 - (23x + 2) / (x + 4)
So, the simplified form of (20x^2 - 12x + 8) ÷ (2x + 8) is 10 - (23x + 2) / (x + 4)
LOL I am sure the problem statement is in error, not your solution.
I appreciate your understanding. If you have any other questions or need further assistance, please feel free to ask.