hi i was reading through the tutorials and havent really understood how to factor out a trinomial like
12x^4 - 5x^2y^2 - 2y^4
what would the 1st few steps be?
Do a substitution:
let v= x^2 and u= y^2
Now you have...
12v^2 - 5uv - 2u^2
(3v -2u)(4v+u) then go back and put the x and y back in...
(3x^2 -2y^2)(4x^2+y^2)
Notice the first paren is the difference between two squares, so it can be factored more..
(xsqrt3 -y sqrt2)(xsqrt3 +y sqrt2)(4x^2+y^2)
To factor out the trinomial 12x^4 - 5x^2y^2 - 2y^4, you can use a substitution to simplify the expression.
1. Let's substitute v = x^2 and u = y^2.
So the expression becomes 12v^2 - 5uv - 2u^2.
2. Next, we need to factor this trinomial.
We can use the factoring method, grouping, or the quadratic formula to solve for the factors.
In this case, let's try factoring by grouping:
Start by multiplying the coefficient of the leading term by the constant term:
12 * (-2) = -24
Find two numbers whose product is -24 and sum is -5 (coefficient of the middle term):
The numbers are -8 and 3 since (-8) * 3 = -24 and (-8) + 3 = -5.
Now we can split the middle term using these two numbers:
12v^2 - 8uv + 3uv - 2u^2.
Factor by grouping:
(12v^2 - 8uv) + (3uv - 2u^2).
4v(3v - 2u) + u(3v - 2u).
Now we can factor out the common binomial:
(3v - 2u)(4v + u).
3. After factoring the trinomial, we now have (3v - 2u)(4v + u).
4. Finally, substitute the original values back in for v and u:
(3x^2 - 2y^2)(4x^2 + y^2).
Now, if you notice that (3x^2 - 2y^2) can be factored further as the difference of two squares, you can continue factoring:
5. The difference of two squares states that a^2 - b^2 = (a + b)(a - b).
From the expression (3x^2 - 2y^2), we can see that 3x^2 is like a^2 and 2y^2 is like b^2.
Now we can factor it further as:
(x√3 - y√2)(x√3 + y√2)(4x^2 + y^2).
Therefore, the factored form of the trinomial 12x^4 - 5x^2y^2 - 2y^4 is (x√3 - y√2)(x√3 + y√2)(4x^2 + y^2).