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Mathematics 19 Online
Kyky232:

I need help finding the solution set

Kyky232:

1 attachment
GalaxyzStarz:

Um, can I see an example? Like, have you done one before?

GalaxyzStarz:

I have to see what the final answer should look like.

Kyky232:

uhm one second

Kyky232:

1 attachment
Kyky232:

the answer was (−3−√7,−3,–√7,6)

GalaxyzStarz:

Okay, I see.

GalaxyzStarz:

So, first, do you know how to factor the equation?

Kyky232:

n o p e :)

GalaxyzStarz:

... Well, you need to learn to start that before anything...

Kyky232:

i just started at my school and we werent here yet

Kyky232:

at my old school

GalaxyzStarz:

Uhm, okay. Do you know basic factoring? Like | \[x^2+10x+25=0\]

Kyky232:

ik how to factor like.. foil

Kyky232:

(x+5)(x+5)=0 if thats what you mean

GalaxyzStarz:

If a polynomial function has integer coefficients, then every rational zero will have the form P/Q where p is a factor of the constant and q is a factor of the leading coefficient. Do you understand that?

Kyky232:

sorta

Kyky232:

so like

Kyky232:

25/ 10?

GalaxyzStarz:

Well, like. \[p=\pm1, \pm 2\] \[q=\pm1\]

Kyky232:

wait how'd you get that

Kyky232:

that makes no sense

GalaxyzStarz:

Graph it.

Kyky232:

OOOOHHHH

GalaxyzStarz:

Substitute 2 and simplify the expression. In this case, the expression is equal to 0 so, 2 is a root of the polynomial.

GalaxyzStarz:

|So, we substitute 2 into the polynomial. | \[2^3-5\times2+2=0\]

surjithayer:

the answer was (−3−√7,−3+√7,6)

surjithayer:

(x+5)(x+5)=0 x=-5,-5 -5 is a repeated zero of the polynomial

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