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since those are your zeros, multiply them up

given intercepts at: a,b,c,d, then:
(x-a)(x-b)(x-c)(x-d)=0 will give you your basic equation

then determine that value of that basic for x=-1 and see what you have to scale by to get it to -6

decimals are fine for factors; but can be written in other ways

the issue is in the scalar; most likely its a 2 to account for that

what do you mean by scalar?

x^2 -1 has the same intercepts as 2(x^2-1) but the graph hits a new set of points everywhere else

i see, with the 2x+1 part youve actually doubled it so you have to divide it back out for the "a"

yeah that's the problem :(

at any rate, your setup was fine.

i don't know whether i should use x+0.5 or 2x+1

2(x+.5) = (2x+1) so its really just a matter of what the author of your text has decided to go with

2x+1 = 0 when x=-.5
x+.5 = 0 when x = -.5

their choice to double it to begin with accounts for a difference in the scalars you choose for "a"

but after everything is expanded out, it should be the same

yours
\[x^4+0.5 x^3-7. x^2+2.5 x+3\]
theirs
\[x^4+x^3/2-7 x^2+(5 x)/2+3\]

oh kk thanks!