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Mathematics 7 Online
OpenStudy (anonymous):

find the value of k so that the remainder of (x^3+5x^2-4x+k)/(x+5) is 0

OpenStudy (freckles):

you are dividing by x+5 so x+5=0 when x=-5 so plug in -5 into the top polynomial and set it equal to 0 that is if f(x)=x^3+5x^2-4x+k then f(-5)=0 <--solve this equation for k

OpenStudy (anonymous):

5=k? 0=k? @freckles

OpenStudy (freckles):

what is f(-5)?

OpenStudy (freckles):

replace the x's in x^3+5x^2-4x+k with (-5)'s

OpenStudy (freckles):

that is called f(-5) then we know that should be equal to 0 since we are given f(-5)=0

OpenStudy (anonymous):

so k=0? @freckles

OpenStudy (freckles):

really don't see how you get k=0

OpenStudy (freckles):

\[(-5)^3+5(-5)^2-4(-5)+k=0 \] I don't think this will give you k=0

OpenStudy (anonymous):

wait k=45?

OpenStudy (freckles):

go through it slowly

OpenStudy (freckles):

(-5)^3=-125 5(-5)^2=125 -4(-5)=20

OpenStudy (freckles):

-125+125+20+k=0

OpenStudy (anonymous):

so many -20

OpenStudy (freckles):

yeah

OpenStudy (anonymous):

so its -20

OpenStudy (anonymous):

@freckles

OpenStudy (freckles):

yes?

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