Perform the division.
1/2 =.5
Complete the square.
We have a clue to what it will look like Hint: the top should have (x+8)
a. x + 9 - 5/x+8 b. x + 9 + 5/x+8 c. x + 10 d. x+9/x+8
It really doesn't matter what the answer key is. You can get the right answer by doing what I said.
Unless you dn't know how to solve it?
Yeah i am clueless.
(x+8) (x+8) x(x+8) (x+8) x^2 finished x(x+8)+9(x+8) 17x finished x(x+8)+9(x+8) -5 the term 67 is finished & -5 is remainder (x+9)-5/(x+8) Did you get the above method?
oh my gosh i think i do! Thank you!
No matter how complex the division is it is the simplest procedure to divide write the equation in the terms of divisor as many times as it is the highest degree term in equation.. keep some spaces in between.. now start consuming each term of the polynomial one by one like the one i did above.
First of all, you should know that we can write \[a^2+2ab+b^2 = (a+b)^2\] Now as @Romero said, our aim should be to cancel the denominator with numerator. For that we need to get (x+8)^2 So what is (x+8)^2? It is \[x^2 + 2(x)(8)+8^2\] \[x^2+16x+64\] So basically you should be looking to get x^2+16x+64. So how do we do that? In the question we have in the numerator \[x^2+17x+67\] which we should write as \[x^2+17x+67=(x^2+16x+64)+(x+3)\] Now we can write \[x^2+16x+64= (x+8)^2\] Therefore we finally get \[x^2+17x+67=(x+8)^2+(x+3)\] Now substituting this in our question, we get \[\frac{(x+8)^2+(x+3)}{x+8}=\frac{(x+8)^2}{x+8}+\frac{x+3}{x+8}\] \[=(x+8)+\frac{x+3}{x+8}\]
Now we can add 5 and subtract 5,so we get \[=(x+8)+\frac{x+3+5}{x+8}+\frac{-5}{x+8}\] \[=(x+8)+1+\frac{-5}{x+8}\] \[=(x+9)-\frac{5}{x+8}\] I hope you understoodknow. The reason I provided you with full answer is to make you understand how to do these kinds of sums in the future :)
There is a long division method too for solving this sum. Let me also show that to you below
|dw:1336679055192:dw| So we can write the final answer as \[=(x+8)+1+\frac{-5}{x+8}\] I haven't gone into explanation of this method. You can always learn it from here--> http://www.purplemath.com/modules/polydiv2.htm
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