# Can we factorize every equation......

Like $$3p^{2} + 5pq - 2q^{2}$$ Can anyone help to factorize these type of equations involving two variables and quadratic.........

Note by Kiran Patel
5 years, 10 months ago

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(3p - q)(p +2q)

- 5 years, 10 months ago

thanxxxx...

- 5 years, 10 months ago

3p^2 + 6pq - pq - 2q^2 3p(p+2q) -q(p+2q) (p+2q)(3p-q)..... ........... in this way it can be solved

- 5 years, 10 months ago

What you have to do is start by finding the coefficients on p. Since it's a 3, you know it will look like (3p q)(p q) where you have to fill in the rest of the numbers. Now, you have to guess and check where to put the 2, and it will have a - either in front of it or in the other factor. You have to fumble around this one, and you will get (3p-q)(p-2q).

Another method, a bit more complicated, works if you don't want to guess and check, like if you have a number with a lot of factors like 12 or 24. Treat the q's as constants, and look at the expression as a quadratic in p. Basically, 3p^2+(5q)p+(-2q^2). You then apply the quadratic formula with a=3, b=5q, and c=-2q^2. Doing this, you get (-5q p/m 7q)/6, which gets you -2q and q/3 depending on the p/m. These, if you remember, are the roots of the equation. To factor, you take the variable (p here) and subtract from that the roots you found, and multiply the two expressions. (p- (-2q))(p-q/3)=(p+2q)(p-q/3) (just like how you would normally do it). Now, you just have to multiply by a number to make sure the coefficients match up. Multiply by 3: 3(p+2q)(p-q/3)=(p+2q)(3p-q) (we could put it on the other factor, but then we'd still have that fraction).

- 5 years, 10 months ago

thanx..

- 5 years, 10 months ago

its simply a quadratic equation involving two variables

- 5 years, 10 months ago

To answer the general question, you may not always be able to factorize an algebraic expression (apart from the trivial factorization). See for example Factorization - Test Yourself 3.

Staff - 5 years, 10 months ago

it is simple you treat p only as variable. in the quadratic equation ax'2+bx+c, rewrite b so as the product of two integers is equal to aXc and their sum or difference should be equal to b , sum or difference is depends up on the signs of the a and c. after that bringing out the necessary common terms, we can write them product of two factors. 32=6, 61=6, 6-1=5, 3p'2+6pq-pq-2q*2=3p(p+2q)-q(p+2q) = (p+2q)(3p-q)

this method has some limitations that when there exists integer factors only.

- 5 years, 10 months ago

32=6, 61=6, 6--1=5

- 5 years, 10 months ago

You can divide the whole equation by \p or \q assuming they are not equal to 0, then the question becomes a problem of solving a quadratic equation.

- 5 years, 9 months ago

3p2 + 5pq-2q2

=> 3p2 + 6pq-pq - 2q2

=> 3p(p+2q) -q(p+2q)

=> (3p-q)(p+2q) = ans

- 5 years, 9 months ago