Question #279504

P(x)=6x3+5x2-2x-1


1
Expert's answer
2021-12-15T00:51:31-0500
P(x)=6x3+5x22x1P(x)=6x^3+5x^2-2x-1

=6x3+6x2x2xx1=6x^3+6x^2-x^2-x-x-1

=6x2(x+1)x(x+1)(x+1)=6x^2(x+1)-x(x+1)-(x+1)

=(x+1)(6x2x1)=(x+1)(6x^2-x-1)

Let 6x2x1=0.6x^2-x-1=0.


D=(1)24(6)(1)=25D=(-1)^2-4(6)(-1)=25

x=1±252(6)=1±512x=\dfrac{1\pm\sqrt{25}}{2(6)}=\dfrac{1\pm5}{12}

x1=1512=13,x2=1+512=12x_1=\dfrac{1-5}{12}=-\dfrac{1}{3}, x_2=\dfrac{1+5}{12}=\dfrac{1}{2}

Then


6x2x1=6(x+13)(x12)6x^2-x-1=6(x+\dfrac{1}{3})(x-\dfrac{1}{2})

=(3x+1)(2x2)=(3x+1)(2x-2)

P(x)=6x3+5x22x1P(x)=6x^3+5x^2-2x-1

=(x+1)(3x+1)(2x2)=(x+1)(3x+1)(2x-2)

Zeros: 1,13,12-1, -\dfrac{1}{3}, \dfrac{1}{2}



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