Factorise 4x(2)y + 12xy(2)?

Okay for starters you know you can divide everything 4 because 4 goes into 12, 3 times.

so now it looks like this: 4 [1x(2)y + 3 xy(2)]

Now you can pull out 1 x and 1 y because both parts have at least one x and y.

Your final answer looks like this: 4xy [1y+ 3y]

Take out the common factor between the two then use that outside the bracket and what's left inside

Common between the two is 4xy, so the answer will give you 4xy(a+b)

where a will be what is left over from 4x(2)y and b will be left from 12xy(2)

the ideal format ought to be (?x +/- n)(?x +/- n) ? & n are the two numbers. each and every of the values interior the 1st set of brackets is elevated by utilising each and every fee interior the 2nd set of brackets. this could provide: ?x^2 +/- ?nx +/- n?x +/- nn Like values are then extra. the place the n? or ?n seems, it in basic terms ability that the two numbers that they characterize are elevated. n?x or ?nx ability the two numbers are elevated yet have the x following them. The final selection on the equation which you're asked to factorize, is produced by utilising multiplying the two n values. as a result +sixteen. So which 2 numbers produce +sixteen? 0 & +sixteen, +2 & +8, +4 & +4. in spite of the incontrovertible fact that, because of the fact the middle fee is a unfavorable, this means that the two between the n values are unfavorable. because of the fact including 2 unfavorable values produces a unfavorable, mutually as multiplying 2 unfavorable values will provide a great! that's the only way that the middle and end numbers could be the two plus and minus (i think of!) So 0 & -sixteen, -2 & -8, -4 & -4 will additionally provide a +sixteen very final fee. The -16x fee is calculated by utilising including the ?nx fee and the n?x fee. they are, after all, the two x to the skill a million values. whilst the two ?x values are elevated, you get an x to the skill 2 result. This fee comes first interior the equation you're asked to factorize. So, we'd desire to discover 2 numbers that as quickly as elevated mutually will provide +sixteen and whilst extra mutually will provide -sixteen. i could advise sixteen & 0. Or 0 & sixteen. So, (4x - sixteen)(x - 0) Will provide 4x^2 - 16x + sixteen in spite of the incontrovertible fact that, I easily have concluded that different areas of the international try this component otherwise; so this would possibly not be what you're finding for!

look at the terms for each of their components looking for common factors

here looking at the numbers ( 4 & 12 ) 4 is a common factor - leaving 3 in the second term

x is a common factor leaving an x in the first term( from x^2 ) but no x is left in the second term

similarly y is a common factor leaving a y in the second term but none in the first

so you get 4(x)^2y + 12x(y)^2 = 4xy ( x + 3y )

as your trying to put the common terms on the outside of the bracket. 4, x and y are all present in both terms, and just go on from there lol.

4xy(x+3y)

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4xy(x + 3y)

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