Question

Would someone like to help me with these Algebra 1 problmes?

3b^2 + 13b + 4 over b + 4

x^2 + 8x + 15 times x^2 - 16
_______________Times___________
x - 4 2x + 6



x^2 + 9x + 20 divided x + 4
_______________divided________
x^2 - 25 x-4

Thank you !

Answer 1

The first thing you should try is factoring. What are the factors of x^2-16?

Answer 2

\[\LARGE {3b^2+13b+4\over b+4}={\left(b+\frac13\right)(b+4)\over b+4 }=\]
\[\LARGE {\left(b+\frac13\right)\cancel{(b+4)}\over \cancel{b+4 }}\] ...

Answer 3

second one..

\[\LARGE {x^2+8x+15 \over x-4}\cdot {x^2-16\over 2x+6 }=\]
\[\LARGE {(x+3)(x+5) \over x-4}\cdot {x^2-4^2\over 2(x+3) }=\]
\[\LARGE {\cancel{(x+3)}(x+5) \over x-4}\cdot {(x-4)(x+4)\over 2\cancel{(x+3)} }=\]
\[\LARGE {(x+5) \over \cancel{x-4}}\cdot {\cancel{(x-4)}(x+4)\over 2 }=\]
\[\LARGE {(x+5)(x+4)\over 2 } \]

Answer 4

and the last one ... what have you tried? :)

Answer 5

I tried the first one I forgot to add What is the undefined value 3b^2 + 13b + 4 / b + 4 so I think the answer is zero since i did in my calculator ..... sorry @Kreshnik

Answer 6

And thank you very much for the second one and sorry for having you do all the work for the second one (:

Answer 7

And the second one I think is x-4 over x-5

Answer 8

Impossible!

Honestly, I'm confused :(
If you have only to simplify, this is it :

\[\LARGE {3b^2+13b+4\over b+4}={\left(b+\frac13\right)(b+4)\over b+4 }=\]
\[ \LARGE {\left(b+\frac13\right)\cancel{(b+4)}\over \cancel{b+4 }}=b+\frac13 \]
unless the expression has =something... in that case it would be:
\[\LARGE b=\text{Something}-\frac13 \]
and the second one...


\[\LARGE {x^2+8x+15 \over x-4}\cdot {x^2-16\over 2x+6 }=\]
\[\LARGE {(x+3)(x+5) \over x-4}\cdot {x^2-4^2\over 2(x+3) }=\]
\[\LARGE {\cancel{(x+3)}(x+5) \over x-4}\cdot {(x-4)(x+4)\over 2\cancel{(x+3)} }=\]
\[\LARGE {(x+5) \over \cancel{x-4}}\cdot {\cancel{(x-4)}(x+4)\over 2 }=\]

unless you don't have to simplify this expression or something , or the given I've taken from your post are not the one which you have in your book\notebook .
and the third one you have is correct...

\[\LARGE \begin{array}{l}3.\\\\\frac{{{x^2} + 9x + 20}}{{{x^2} - 25}}:\frac{{x + 4}}{{x - 4}} = \\\\ = \frac{{\left( \cancel{{x + 5}} \right)\left( \cancel{{x + 4}} \right)}}{{\left( {x - 5} \right)\left( \cancel{{x + 5}} \right)}} \cdot \frac{{x - 4}}{\cancel{{x + 4}}} = \\\\ = \frac{{x - 4}}{{x - 5}}\end{array}\]

.
If this is wrong... with all do respect sir, I'd be delighted to know the correct way of doing these problems. Thanks. @Tutu55

Answer 9

sorry I wan't here before...