Asked by CheezyReezy
Solve the following equation in the real number system. Please show all of your work.
x^4 + 6x^3 + x^2 - 24x - 20 = 0
Help!!!!
x^4 + 6x^3 + x^2 - 24x - 20 = 0
Help!!!!
Answers
Answered by
Mgraph
Let f(x)=x^4+6x^3+x^2-24x-20
Real roots are looking for among
+-1,+-2,+-4,+-5,+-10,+-20
f(-1)=1-6+1+24-20=0
Divide f(x) by (x+1):
f(x)=(x+1)(x^3+5x^2-4x-20)
Let g(x)=x^3+5x^2-4x-20
g(-5)=-125+125+20-20=0
Divide g(x) by (x+5):
g(x)=(x+5)(x^2-4)
f(x)=(x+1)(x+5)(x+2)(x-2)
Real roots are looking for among
+-1,+-2,+-4,+-5,+-10,+-20
f(-1)=1-6+1+24-20=0
Divide f(x) by (x+1):
f(x)=(x+1)(x^3+5x^2-4x-20)
Let g(x)=x^3+5x^2-4x-20
g(-5)=-125+125+20-20=0
Divide g(x) by (x+5):
g(x)=(x+5)(x^2-4)
f(x)=(x+1)(x+5)(x+2)(x-2)
Answered by
billy
what is 2/5 in the real number system
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