Find all rational zeros of the polynomial

Mathematics

1 answer:

rjkz [21]3 years ago

4 0

$x^3+x^2-5x+3=x^3+3x^2-2x^2-6x+x+3\\\\=(x^3-2x^2+x)+(3x^2-6x+3)=x(x^2-2x+1)+3(x^2-2x+1)\\\\=(x^2-2x+1)(x+3)=\underbrace{(x^2-2(x)(1)+1^2)}_{(a-b)^2=a^2-2ab+b^2}(x+3)=(x-1)^2(x+3)\\\\P(x)=0\iff(x-1)^2(x+3)=0\iff(x-1)^2=0\ \vee\ (x+3)=0\\\\x-1=0\ \vee\ x-1=0 \vee\ x+3=0\\\\x=1\ \vee\ x=1\ \vee\ x=-3\\\\Answer:\ \boxed{x_1=1,\ x_2=1,\ x_3=3}$

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Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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