Question

factorise a^3-a-120

Answer 1

heya friend ❤❤

here's ur answer dear ⬇⬇⬇

First find out the factors of 120. They are 1, 2, 3, 4, 5,6,8,10,12,15,20,24,30,40,60,120 and the negative of these values i.e -1, -2 and so on.

Now you have to see, for which of the numbers given above, a^3-a-120 becomes 0. Since there are negative signs, you can only check the positive factors.

We see that if a=5, a^3-a-120=125-5-120=0. So

a-5 is a factor of a^3-a-120. So divide a^3-a-120 by a-5. The quotient is a^2+5a+24. You cannot break a^2+5a+24 further into simpler terms.

Thus, a^3-a-120=(a-5)(a^2+5a+24)



hope it helps :-)

#thank you

Answer 2

Hey friend here is your answer