SOLUTION: Factorise: abc+5a^2c x^2+6x+9 p^2-q^2 x^2+x-6

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: Factorise: abc+5a^2c x^2+6x+9 p^2-q^2 x^2+x-6      Log On


   

Question 56701: Factorise:
abc+5a^2c
x^2+6x+9
p^2-q^2
x^2+x-6
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Factor:
abc%2B5a%5E2c
ac%28b%29%2Bac%285a%29
highlight%28ac%28b%2B5a%29%29
:
x%5E2%2B6x%2B9
(x____)(x____)
we need to fill the blanks with two integers that multiply to give us:+9 and add to give us:+6
3*3=9 and3+3=6
%28x%2B3%29%28x%2B3%29 This is a perfect square trinomial:highlight%28a%5E2%2B2ab%2Bb%5E2=%28a%2Bb%29%5E2%29
highlight%28%28x%2B3%29%5E2%29
:
p%5E2-q%5E2 This is the difference of two squares:highlight%28a%5E2-b%5E2=%28a%2Bb%29%28a-b%29%29
highlight%28%28p%2Bq%29%28p-q%29%29
:
x%5E2%2Bx-6
(x___)(x___)
We need to fill the blanks with two integers that multiply to give us:-6, but add to give us:+1
3*-2=-6
3-2=+1
highlight%28%28x%2B3%29%28x-2%29%29
Happy Calculating!!!