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x^4-7x^3-12x^2+176x-320 I have to give the exact values list multiple zeros as necessary so i just need to know the probably 4 zeros of this prob
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x=4 is a zero
so x^4-7x^3-12x^2+176x-320=(x-4)(x^3-3x-24x+80)=0
there is more then 1 zero isnt there
(x-4)(x^3-3x^2-24x+80)=0*
yes there are 3 more
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nice!
x=4 is again another zero
(x-4)(x-4)(x^2+x-20)=0
now we have 2 more to find
+5 - 4
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-5,4
oo ya i didnt factor them to 0
because x^2+x-20=(x+5)(x-4)
thank you !
so 4 multiplicity 3 and 5
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oops -5
np
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