You make a table of x vs. y values. That is, you substitute some value for "x", and calculate the corresponding value for "y". Then you graph the points. The equations you give are straight lines, so two points are sufficient - in theory. It is recommended to include three points, as an additional check, in case you do something wrong.
2x2 + 11x + 15 = (x + 3)(2x + 5).
-2x plus 3y equals 1
Yes
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
2x2 + 3x - 20 = (x + 4)(2x - 5).
2x2 + 11x + 15 = (x + 3)(2x + 5).
2x2=4x4=16
-2x plus 3y equals 1
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Yes
x = (-1 ± √11) / 2
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
2x2 + 3x - 20 = (x + 4)(2x - 5).
2x2 + 10 = 9x 2x2 - 9x + 10 = 0 (x - 2)(2x - 5) = 0
2x2 + 10 - 9x = 0 or 2x2 - 9x + 10 = 0 So (x - 2)*(2x - 5) = 0
7
No.