Find the equation of the line that passes through (1,-1) and is parallel to 2x+y−3=0. Leave your answer in the form y=mx+c

y= -2x + 1

Step-by-step explanation:

We can formulate the slope equation of that line by

S= Y-Y1/X-X1

By the given statement "parallel", the slope of the function "y=3-2x" shall then equals the first line -->1

to conclude, we will use the first point at x=1 ( Like the given point of the line) and additional point at x=0...

y(1)=3-2(1)=1, Hence, (1,1)

y(0)=3-2(0)=3, Hence, (0,3)

By the fact 1

then we shall conclude with simple linear equation as follow

m=y2-y1/x2-x1 = 1-3/1-0=-2

The general formula of the Required function is y=mx+c

In fact the line passes through the point (1,-1)

Hence y in x=1 equals -1

-1= -2(1)+C

C= -1+2=1

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0