Factorise x^3 + 2x^2 - 23x-60
Answers
Step-by-step explanation:
eNotes Home
Homework Help
Study Guides
Texts
Teachers▻
Sign In
Join
rowseNotessearch
Search for any book or any question
Math Questions and Answers
MENU
What are the roots of x^3 – 2x^2 – 23x + 60?
print Print document PDF list Cite
Expert Answers info
TUSHAR CHANDRA eNotes educator | CERTIFIED EDUCATOR
We have to find the roots of : x^3 – 2x^2 – 23x + 60
x^3 – 2x^2 – 23x + 60 = 0
As the equation has a highest power of x, we will have 3 roots.
From the numeric term equal to 60 we know that the product of the roots is 60, substituting values 1,-1, 2, -2 and 3, we get that 3 is a root. So now we have to factor out x – 3, the other roots have a product of 20.
x^3 – 2x^2 – 23x + 60 = 0
(x^3 + x^2 – 20x – 3x^2 – 3x + 60) = 0
(x – 3)(x^2 + x – 20) =0
(x – 3)( x^2 + 5x – 4x – 20) = 0
We can factorize the quadratic term as 5* -4 = -20 and 5 - 4 = 1
(x – 3)( x(x + 5) – 4(x + 5)) = 0
(x – 3)(x – 4)(x + 5) = 0
This gives the roots of x^3 – 2x^2 – 23x + 60 as x = 3 , x = 4 and x = -5.
The required roots are
x = 3 , x = 4 and x = -5.
Answer:
,
Step-by-step explanation:
as it's a cubic equation the factors of the equation will be 3 small brackets therefore further you solve you get the answer of x.
