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Factorise this equation.

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2 Answers

This best works by combining the terms. Use powerful words "let us notice that...". Those are 4 very powerful words in math. 

Let us notice that we have mn factor in the first term and either m or n in the second or third term, respectively. Naturally, let us combine first and third term together and second and fourth terms are to be combined too.

6mn+4m+3n+2=(6mn+3n)+(4m+2);

Now let us factor 3n out of the first parentheses. We will obtain:

6mn+4m+3n+2=3n(2m+1)+2(2m+1), here we also factored 2 out of the second parentheses. Now let us notice that (2m+1) is a common factor here and it can be therefore factored out. So we will get the following:

6mn+4m+3n+2=(2m+1)(3n+2) Factorization is done.

You would get the same result if you combine the terms differently, say 1st and 2nd together and 3rd and 4th together. Try it yourself the same way I did and check  that you will get the same answer.

Hope it helps. This method may prove very useful for solving some equations, like cubic one, for example.

x^3-6*x^2-3x+18=0 is one example. Try to solve it. I will help if you need the assistance.

Work from the back end of the expression: (3n+2) = 1*(3n+2) and assume that 1*(3n+2) is somehow a factor of the first two terms in the expression.  That is ASSUME that (3n) can give me 6mn if I multiply it by 2m, and so 2m(3n+2) gives 6mn+4m.  Good so far.  Now UN-distribute the factors as follows (2m)(3n+2) +1(3n+2) = 6mn + 4m + 3n + 2.