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If you have the equation y=3x2+6x then you can factor it down smaller like so:

y=3x(x+2)

This is possible because there is an 'x' in both parts of the equation. You can also do it backwards, because 3x*x=3x2 and 3x*2=6x.

If you are talking about a regular problem, such as 3+4(3-7) then you'd do the stuff in parentheses first. Remember the order: PEMDAS. Parentheses, Exponents, Multiply, Divide, Add, Subtract.

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12y ago
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12y ago

There are three ways to factor with different variables.

- using the distributive property

-using the additive inverse property

-factoring by grouping

Grouping

Most people believe that factoring by grouping is the easiest

factor 4ab+8b+3a+6

= (4ab+8b)+(3a+6) group terms together with common factors

=4b(a+2) + 3(a+2) factor GCF from each grouping

= (a+2) (4b+3) use distributive property

Distributive

factor 12a2+6

First, find GCF of 12a and 16a by factoring each number

12a= 2*2*3*a*a

16a= 2*2*2*2*a

GCF= 2*2*a or 4a (like factored terms)

Now, write each term as the product of the GCF and its remainning factors. Then use the Distributive property to factor out the GCF

12a2+16a= 4a(3*a) + 4a(2*2)

= 4a(3a) + 4a(4)

= 4a(3a+4)

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11y ago

Just like any other quadratic/polynomial. Are there any common factors? Use GCF.

Do all the variables follow rules for a quadratic? Then consider a dummy variable to replace them. At the end of the problem you must do reverse substitution.

Ex: 5x2y2z2 + 15xyz +10 <-- GCF is 5

= 5(x2y2z2 + 3xyz +2) <--notice that x2y2z2 + (xyz)2 and sub. A = xyz so A2 = x2y2z2

= 5(A2 + 3A + 2) = 5(A+1)(A+2) <-- now must reverse substitute xyz = A

= 5(xyz +1)(xyz+2) <-- this is your answer

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10y ago

Use the quadratic formula to find the roots in the complex field, and use the result to factorise the trinomial.

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Q: How do you factor quadratics with 3 variables?
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