Question

how to factor trinomials that are the product of simular binomials ?

Answer #1

10t^2 +t -3

We know that the brackets are going to be setup like such:

(?t+/-?)(?t+/-?)

Factors of 10 will go in the first ? of each bracket. Factors of 3 will go in the second ? of each bracket.

NOTE since 3 is negative, we are going to have one bracket positive, and the other bracket negative:

(?t+?)(?t-?)

Factors of 10: 10 and 1, 2 and 5

Factors of 3: 1 and 3

NOW we're going to sub in the numbers, and hopefully when we multiply it out, we wil get the middle term; 1t. START with the groups of numbers that are closest together

(2t+3)(5t-1) = 10t^2 -3 +15t -2t

= 10t^2 +13t -3

Not exactly., so let's try switching the numbers around:

(5t+3)(2t-1) = 10t^2 -3 +6t -5t

= 10t^2 +1t -3

AHA! That's the solution set

10t^2 +t -3 is factored as

(5t+3)(2t-1)

6n^2 -13n +6

once again, same thing, except this time, we can see that while the last number is positive, the middle number is NEGATIVE; that means that both brackets will be negative:

(?n-?)(?n-?)

factors of 6: 6 and 1; 2 and 2

remember; we're trying to get -13n:

(2n-3)(3n-2) = 6n^2 +6 -9n -4n

= 6n^2 -13n +6

Which is our original expression

so 6n^2 -13n+6 is factored as

(2n-3)(3n+2)

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