Question

anyone familiar with polynomial functions?

Answer 1

May I help?

Answer 2

what do you need clarification on?

Answer 3

OK, what is your specific question?

Answer 4

find p(4) and p(-2) for each function.
p(x)=2-x

Answer 5

OK, so p(x)=2-x
p(4) = 2-(4) = -2
p(-2) = 2-(-2) = 4

Answer 6

Just plug in the number inside p(x) into the function.

Answer 7

thank you, i have one more problem,
if p(x)=3x^2-2x+5 and r(-x)=x^3 +x+1
find 2[p(x+4)]

Answer 8

What is the use of r(-x) here when we have to find 2[p(x+4)]? Anyways

2[p(x+4)]=2[3(x+4)^2-2(x+4)+5]=2[3(x+4)^2 -2x-8+5] = 2[3(x^2+8x+16)-2x-3]=2[3x^2+24x+48-2x-3]=6x^2+44x+90

\[6x^2+44x+90\]

Answer 9

solve polynomial equations\[x-6\sqrt{x}=7\]

Answer 10

can anyone help me solve polynomial functions

Answer 11

put \(\sqrt x=y\)
then x=y^2
so, you have a quadratic equation,
\(y^2-6y-7=0\)
can you solve this by factorization ??

Answer 12

wait... why did you set \[\sqrt{x}=y\]?

Answer 13

so that i get a quadratic equation, which has standard ways to solve.

Answer 14

but where'd the y come from?

Answer 15

its a substitution
after you find 'y' you re-substitute y =\(\sqrt x\) to find the required values of x

Answer 16

but there's no y in the original equation.

Answer 17

thats what a substitution means.
you replace the function of the required variable with another variable so that the equation can be solved easily.

Answer 18

so just put "y" in place of zero?

Answer 19

y in place of 0 ?? i didn't say that
put 'y' in place of \(\sqrt x\)
so, x becomes y^2

Answer 20

i don't understand

Answer 21

which part ?

Answer 22

why are you putting y in place of \[\sqrt{x}\]? It doesn't make sense. I know it's to make it a quadratic equation but i don't get where the y is coming from.

Answer 23

'y' is just another variable. its coming from your mind to simplify things and your life.
you can use any other variable in place of 'y' like 'a', 's', 'w', 'z' or even "BrainWest' :P

Answer 24

@BrianWest Hilarious!

Answer 25

so, you wouldn't try to factor out an 'x' or anything?

Answer 26

since 'x' cannot be factored , no.
after you find 'y' you re-substitute y =√ x to find the required values of x

Answer 27

where did you get the y

Answer 28

i am not answering that again, accept it or leave it.

Answer 29

ok thank you

Answer 30

here is another on please..
\[x ^{3}+343=0\]

Answer 31

thats has 2 complex roots and 1 real root. you need to get the real root only, right ?
\(\large x^3=-343 \ \large \x= \sqrt[3]{343}=...?\)

Answer 32

actually both, thank you

Answer 33

so then you'd have to factor out x+7 form x^3+343 =0 by long division or synthetic division to get a quadratic equation which you can solve by quadratic formula.