x³-3x+2 factorise this some
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f(x)=x^3-3x+2
f(1)=(1)^3-3(1)+2=1-3+2=0
therefore,x-1 is the factor.
by division,
x^3-3x+2
=(x-1)(x^2+x-2)
=(x-1)(x^2-x+2x-2)
=(x-1)(x(x-1)+2(x-1))
=(x-1)^2(x+2)
roots={1,1,-2} double root at 1
f(1)=(1)^3-3(1)+2=1-3+2=0
therefore,x-1 is the factor.
by division,
x^3-3x+2
=(x-1)(x^2+x-2)
=(x-1)(x^2-x+2x-2)
=(x-1)(x(x-1)+2(x-1))
=(x-1)^2(x+2)
roots={1,1,-2} double root at 1
dhonisuresh0703:
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