X^3-4x^2+3x+2=0 find all the real values of x

Step-by-step explanation:

The factor of the polynomial is (x-2)

We know that dividing a third degree polynomial by a first degree factor will yield a quadratic of the form ax²+bx+c and possibly a remainder r,except that in this case r=0 since x-2 is a factor

We therefore equate

x³-4x²+3x+2=(x-2)(ax²+bx+c)+r

x³-4x²+3x+2=ax³+bx²+cx-2ax²-2bx-2c+r

We group terms on the RHS

x³-4x²+3x+2=ax³+x²(b-2a)+x(c-2b)-2c+r

By equating the terms on both sides,

ax³=x³,a=1

Next,

b-2a=-4

b-2(1)=-4

b-2=-4

b=-4+2

b=-2

Next,

c-2b=3

c-2(-2)=3

c+4=3

c=-1

Now we piece together our quotient quadratic using a=1,b=-2,c=-1 to obtain x²-2x-1

We now find the zeros of x²-2x-1

which are 1+√2 and 1-√2

The values of x are therefore 2,1+√2, and 1-√2

Note that I used this method in lieu of the famous polynomial division approach to teach someone also ;)

2, 1 +/- sqrt2

or 2, 2.414, -0.414 (correct to the nearest thousandth).

Step-by-step explanation:

Let x = 2: then

f(2) = 8 - 16 + 6 + 2 = 0

So x = 2 is one  of the zeros.

So one factor of the function is x-2.

If we divide  the function by  (x - 2) we get:

x-2)X^3-4x^2+3x+2(x^2-2x-1

     x^3- 2x^2

            -2x^2+3x

            -2x^2+4x

                       -x+2

                       -x+2

So x^2-2x-1= 0

x = [-(-2 +/- sqrt(4 - 4*-1)] / 2

x = (2 +/- sqrt8) / 2

= 1 +/-sqrt8/2

= 1 +/- sqrt2

step-by-step explanation: