Question

PreCalc & Trig help please! Question is down below in the picture, I have no idea where to start. I do know how to do synthetic division though. Need to show all work!

Answer 1

http://www.purplemath.com/modules/synthdiv.htm

Answer 2

I know how to do synthetic division, just dont know how to do this problem.

Answer 3

x=2/7 is factor of a polynomial,this mean that x=2/7 satisfies the equation one gets from the polynomial.this also means that (x-2/7) is a factor of the polynomial.now synthetically divide the given polynomial by (x-2/7).the quotient will be quadratic polynomial.factor it,and you get the other two roots

Answer 4

okay, ill give that a try, thanks.

Answer 5

yeah.. have no idea what to do.

Answer 6

if you know how to do synthD, then where are you having problems?

Answer 7

as i see it, either you know how to do synthD, and the problem is then self explanatory; OR you dont know how to do synthD in which case the problem may present itself as a bit confusing ....

but you say you know how to do synthD ....

Answer 8

We've never had to do one with a fraction.
As well as we've never had to do one worded this way

Answer 9

:) fractions are just numbers ... let me see how you set it up to begin with

Answer 10

i have my own method that i think is more intuitive a process than what ive seen in the textbooks

Answer 11

wait, hold on i wrote something on my paper wrong..

Answer 12

okay, that's why i have major issues, i wrote down 12 instead of 126.

so once I get through the division, what are the real solutions? is that like finding the zeros? like, you'd have to simplify the polynomial at the end of the division and factor it, then set each equal to zero to find the real solutions?

Answer 13

yeah, that setup is not very helpful to me, and if a method or setup is not helpful ... then it serves no purpose IMO

what i like to do is set it up so the poly, in full, is the first row

then the second row is for adding

and the last row is for multiplication.


49x^3 -126x^2 +60x -8 <-- poly in full (all the x parts are there)
0 <-- add this row
------------------------
x= ) 49 <-- multiply this row

since x = 2/7, what is 49 * 2/7?

Answer 14

notice that the x= parts is beside the row that it actually interacts with as opposed to someplace up top out of a line of sight

Answer 15

14

Answer 16

49x^3 -126x^2 +60x -8 <-- poly in full (all the x parts are there)
0 14 <-- add this row
------------------------
2/7 ) 49 -112 <-- multiply this row

since x = 2/7, what is -112 * 2/7?

Answer 17

-32

Answer 18

49x^3 -126x^2 +60x -8 <-- poly in full (all the x parts are there)
0 14 -32 <-- add this row
------------------------
2/7 ) 49 -112 28 <-- multiply this row

and again, 28* 2/7?

Answer 19

does this setup make sense to you? or do you like the textbook method better?

Answer 20

either way it's kind of the same thing, just set up differently

Answer 21

by keeping the x parts up top intact, it gives me a guide to go by to form the quotient poly, just reduce the exponents by 1

49x^3 -126x^2 +60x -8 <-- poly in full (all the x parts are there)
0 14 -32 8 <-- add this row
------------------------
2/7 ) 49 -112 28 0 <-- multiply this row
x^2 x^1 x^0

so our quotient poly is:

49x^2 -112x + 28

(x-2/7) (49x^2 -112x + 28)

(x-2/7) 7 (7x^2 -16x + 4) is our factors so far

Answer 22

correct, i've gotten to there.

Answer 23

so just factor the quad and you should be done

Answer 24

except that im awful at factoring. :/

Answer 25

no need to factor, youve got the quadratic formula to plug and play with to determine the values of x in a quadratic equation

Answer 26

7x^2 -16x + 4\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{16\pm\sqrt{16^2-4(28)}}{14}\]
if the under radical is negative, then there are no more real roots

Answer 27

what is that, 256 - 112? so it looks real to me :)

Answer 28

\[16\div14 (+-) 12\div14\]
Then it's 2 and 2/7

Answer 29

i dont get what the answer is supposed to be lol

Answer 30

yes, the roots are then x=2 and x=2/7 from the quad.

so set up 2 factors that zero out: (x-2) = 0 when x = 2
(x-2/7) = 0 when x = 2/7

this gives us a finally of:\[f(x)=7(x-\frac27)(x-\frac27)(x-2)\]simplify to your hearts content

Answer 31

So the real solutions are 2 and 2/7 or -2 and -2/7?

Answer 32

i droped a 7 along the way ... ohwell.

no, there are 3 real solutions ... 2/7 is just a multiple by a factor of 2

Answer 33

roots: 2, 2/7, 2/7

Answer 34

and then the whole thing is scaled by 49

Answer 35

Yup, you lost me. lol

Answer 36

or if you perfer:\[(7x-2)(7x-2)(x-2)\]:)

Answer 37

still lost. lol, thanks, ill try to figure it out