2 years ago

PLEASE HELP ME please

Mathematics

1 answer:

Lena [83]2 years ago

7 0

Answer:

(-1, 1/10) and (-2, 1/100)

I can't really explain because I just put it into my graphing calculator and clicked the trace button to find the y value at certain x values. A good website to do that on is Desmos

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2 years ago

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Ratling [72]

Answer:

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Step-by-step explanation:

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2 years ago

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rjkz [21]

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