(x+1)^2 * (x+3)^2 = 0
It must have two double roots, since there are only two zeroes.
Multiply it out as a fourth order polynomial.
What is a quartic function with only the two real zeroes given? x=-1 and x=-3
Would someone please explain
6 answers
Not sure what you mean
not sure if I did it right but is it y=x^4+4x^3+4x^2+4x+3
not even close
take it step by step till you get
x^4+8x^3+22x^2+24x+9
Did you forget to multiply twice by (x+3)? You gotta have a 9 at the end.
Take a look at calc101.com and click on its long multiplication link.
Enter the two expanded squares:
x^2+2x+1
x^2+6x+9
take it step by step till you get
x^4+8x^3+22x^2+24x+9
Did you forget to multiply twice by (x+3)? You gotta have a 9 at the end.
Take a look at calc101.com and click on its long multiplication link.
Enter the two expanded squares:
x^2+2x+1
x^2+6x+9
Steve can you explain what Im doing wrong the only choices I have sre;
y=x^4-4x^3-4x^2-4x-3
y=-x^4+4x^$x^2+4x+3
y=x^4x^3+3x^2+4x-4
y=x^4+4x^3+4x^2+4x+3
y=x^4-4x^3-4x^2-4x-3
y=-x^4+4x^$x^2+4x+3
y=x^4x^3+3x^2+4x-4
y=x^4+4x^3+4x^2+4x+3
x=2 and x=8
1 answer