f(x)= x^5 - x^4 + 5x^3 - 2x^2 -12x +5; [1.3, 1.6]

f(1.3)=

f(1.6)=

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f(x)=x^5-x^4+5x^3-2x^2-12x+5

f(1.3)=3.71293-2.8561+10.985-3.38-15.6+5= -2.13817

f(1.6)=10.48576-6.5536+20.48-5.12-19.2+5=5.09216

Yes... the polynomial has zero between x=1.3 and x=1.6 as f(x) is crossing x-axis between x=1.3 and x=1.6 .

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