Question

the ones with ( ) will show you the answer I choose :)
Simplify (15x2 – 24x + 9) ÷ (3x – 3) = ?
( A. 5x – 3 )
B. 5x + 13 with a –30 remainder
C. 5x – 13 with a –30 remainder
D. 5x + 3
Simplify (2x + 3)(x – 4) = ?
A. 2x2 + 5x + 12
(B. 2x2 + 5x – 12)
C. 2x2 – 5x – 12
D. 2x2 – 5x + 12
What is the result if you divide 18r4s5t6= ?
-3r2st3
A. –6r2s4t3
B. 6r2s4t3
C. –6r2s5t3
D. 6r2s5t3

not sure about this one tho ^^^

Answer 1

Check again the second one.

Answer 2

The first one is good. The last one is the same as the first one but instead of having x's you have r's

Answer 3

hint:
first question:
we have to find the roots of the polynomial 15x^2-24x+9. Do you know how to compute those roots?

Answer 4

or, in other words we have to solve this equation:
15x^2-24x+9=0

Answer 5

no im not smart in algebra never was lol so its why im trying but cant seem to find the answer also sorry for the late reply my wifi is acting up

Answer 6

the roots of that equation are given by the subsequent formula:
\[\Large \begin{gathered}
x = \frac{{12 + \sqrt {{{12}^2} - 15 \times 9} }}{{15}} = ...? \hfill \\
\hfill \\
x = \frac{{12 - \sqrt {{{12}^2} - 15 \times 9} }}{{15}} = ...? \hfill \\
\end{gathered} \]

Answer 7

can you complete the computations above?

Answer 8

honestly, no :(

Answer 9

here are more steps:
\[\Large \begin{gathered}
x = \frac{{12 + \sqrt {{{12}^2} - 15 \times 9} }}{{15}} = \frac{{12 + \sqrt {144 - 135} }}{{15}} = \hfill \\
= \frac{{12 + \sqrt 9 }}{{15}} = \frac{{12 + 3}}{{15}} = \frac{{15}}{{15}} = 1 \hfill \\
\end{gathered} \]
that is the first root

Answer 10

then, the second root is:
\[\Large \begin{gathered}
x = \frac{{12 - \sqrt {{{12}^2} - 15 \times 9} }}{{15}} = .\frac{{12 - \sqrt {144 - 135} }}{{15}} = \hfill \\
= \frac{{12 - \sqrt 9 }}{{15}} = \frac{{12 - 3}}{{15}} = \frac{9}{{15}} = \frac{3}{5} \hfill \\
\end{gathered} \]

Answer 11

so the requested roots are:
\[\begin{gathered}
x = 1 \hfill \\
x = \frac{3}{5} \hfill \\
\end{gathered} \]

Answer 12

now, there is a theorem which states that we can rewrite the polynomial 15x^2-24x+9, as below:
\[\Large 15{x^2} - 24x + 9 = 5\left( {x - \frac{3}{5}} \right)3\left( {x - 1} \right)\]

Answer 13

so your fraction can be rewritten as below:
\[\Large \frac{{15{x^2} - 24x + 9}}{{3x - 3}} = \frac{{5\left( {x - \frac{3}{5}} \right)3\left( {x - 1} \right)}}{{3\left( {x - 1} \right)}} = ...?\]

Answer 14

please simplify the right side, what do you get?

Answer 15

when I simplify will I divide or multi?

Answer 16

hint:

Answer 17

ohh okay so is that what I do with my last question I have up there?

Answer 18

you have to cancel similar terms, as I have made above

Answer 19

yes! The result is:
\[\Large \begin{gathered}
\frac{{15{x^2} - 24x + 9}}{{3x - 3}} = \frac{{5\left( {x - \frac{3}{5}} \right)3\left( {x - 1} \right)}}{{3\left( {x - 1} \right)}} = \hfill \\
\hfill \\
= 5\left( {x - \frac{3}{5}} \right) = 5x - 3 \hfill \\
\end{gathered} \]
so what is the right option?

Answer 20

Simplify (2x + 3)(x – 4) = ?
C. 2x2 – 5x – 12
because
2x*x + 2x*(-4) + 3*x + 3*(-4)

2x^2 -8x + 3x - 12

2x^2 - 5x - 12
am I correct?

Answer 21

your answer is correct!

Answer 22

see for the last question after the one I just got correct can you please make a picture of that like you didfor those ? idk if you know what im talking about

Answer 23

for your last question, we can write:
\[\Large \frac{{18{r^4}{s^5}{t^6}}}{{ - 3{r^2}s{t^3}}} = - \frac{{18}}{3} \times {r^{4 - 2}} \times {s^{5 - 1}} \times {t^{6 - 3}} = ...?\]

Answer 24

where I have applied the property of division of powers with same bases

Answer 25

what do you get?

Answer 26

so its c?

Answer 27

The process seems x2 more complicate than a long division o.o