By splitting the Middle Term, 5X2 + 25X + 4X +20 5X(X+5) + 4(X+5) (5X+4)(X+5)
5x2=10...take 5 two times..5+5=10
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
5x2 = 3 then x2 = 3/5 so that x = sqrt(3/5) = ± 0.7746
5x2 + 20 = 5*(x2 + 4) which cannot be factorised further.
By splitting the Middle Term, 5X2 + 25X + 4X +20 5X(X+5) + 4(X+5) (5X+4)(X+5)
5-8x2-5x2 is equal to -21.
5x2=10...take 5 two times..5+5=10
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
5x2 - 136 = 44 5x2 - 136 + 136 = 44 + 136 5x2 = 180 5x2/5 = 180/5 x2 = 36 √x2 = √36 x = ±6
To factor any term, divide by the LCF (largest common factor) of the term. For 5x2-45, the LCF is 5. 5x2-45 = 5(x2-9) It can, of course, be factored even further to become: 5(x2-9) = 45(x2/9 - 1) but when factoring, we usually don't need to go past 5(x2-9) (for example, you've figured out that x2 = 9, thus x = +/- 3.)
5*2=10 5x2=10 5X2=10
5x2 = 3 then x2 = 3/5 so that x = sqrt(3/5) = ± 0.7746
5x2 + 20 = 5*(x2 + 4) which cannot be factorised further.
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
= 5x2+70-16+9x-2 = 5x2+9x+52 = 5x2+9x1+52 This implies coefficient of degree 1 is 9. Ans.
No.