metalilty

It's a quadratic expression ax^2+bx+c where a=/=1. This is what you do: 1. Take the product of ac. Ignore the signs. 2. Write down the possible pairs of factors of absolute value ac. 3. - If c is positive select the two factors of ac whose sum is equal to b. - If c is negative select the two factors of ac whose difference is equal to b. 4. Write down bx as the sum(or difference) of two factors obtained above. 5. perform grouping 6. write down the factors. 6x^2 + 11x - 10 ac= 60 Factor pairs of ac: 12 and 5; 6 and 10; 1 and 60; 20 and 3; 30 and 2; 15 and 4 Since c is negative we use 15x-4x to replace 11x. 6x^2 + (15x-4x) - 10 = 6x^2 + 15x-4x - 10 Group it: (6x^2 + 15x)-(4x +10) Factor out the common factor: 3x(2x+5)- 2(2x+5) Factor out the common factor again: 2x+5(3x-2) usually written as: (2x+5)(3x-2) As for: 2x^2 + 4x -3 It's unfactorable. This is how you check "factorability" 1.b^2- 4(a) [c] 2. If the result is a perfect square then it's factorable. 1.16- 4(2) (-3)=40 2. √40=2√10 unfactorable. Lets check the previous example: 6x^2 + 11x - 10 1. 121- 4(6) (-10)=361 2. √361=19 factorable.

Simplify the following, giving the result without fractional indices: [(x^2 -1)^2 * √(x+1)]/ (x-1)^3/2 Attempt at solving: There are no common bases to add the indices and no common indices to multiply out the bases so I tried this and got it wrong, please show me where though: [(x-1) (x+1)]^2 * (x+1)^1/2] / (x-1)^3/2 =[(x-1)^2 * (x+1)^2 * (x+1)^1/2] / (x-1)^3/2 multiplied the indices by 2 and got rid of the fractions =[(x-1)^4 * (x+1)^4 * (x+1)] / (x-1)^3 = (x-1) *(x+1)^5 By the book it should be : (x+1)^2* √(x^2-1) Thank You.