21.)P(x0=2.5x^4-2x^2 ---- (22.) P(x)= 12x^3 - 15x^2+ x+1 --- (23.) p(x)=8x^3-6x62-2x+1?

Question

use a graph and the location Principle to find the real zeros of each function. Give approximate values to the nearest tenth, if necessary? please help me solve this!

Answer 1

2.5x^4-2x^2 =0
x^2(2.5x^2 -2) = 0
x = 0,0,sqrt(.8), -sqrt(.8)

12x^3 -15x^2 +x +1 = 0
Approximate solutions are
x = -.2, .35, 1.1

8x^3-6x^2 -2x +1
x = -.43, .32, .87

Answer 2

(2x + a million) (x-a million) ok you will desire to multiply out the brackets, initiate via multiplying the 2x and the x (the 1st determine in each bracket) that gives you you 2(x squared) *sorry, do no longer understand the thank you to do the "squared" image on right here!* then you definately multiply the 2x via the 2nd determine in the different bracket like so: 2x X -a million = -2x that gives you you: 2(x squared) - 2x as much as now then you definately multiply the 2nd determine in the 1st bracket via the 1st determine in the 2nd bracket: a million X x , which basically equals x. as much as now you have 2(x squared) - 2x + x finally multiply the final figures in each bracket: a million x -a million = -a million So the equation is now 2(x squared) - 2x + x -a million or extra basically: 2 (x squared) - x - a million wish that explains it worried with you :-)