Another method for factoring these kinds of quadratic trinomials is called factoring by grouping. Factoring by grouping can be a bit more tedious, and is often not worth the trouble if you can find the correct factors by some quick trial and error. However, it works quite well when the factors are not immediately obvious, such as when you have a very large number of candidate factors. When this happens, the trial and error method becomes very tedious
Factoring by grouping is best demonstrated with a few examples.
Example:
Given: |
5x^{2} + 11x + 2 |
Find the product ac: |
(5)(2) = 10 |
Think of two factors of 10 that add up to 11: |
1 and 10 |
Write the 11x as the sum of 1x and 10x: |
5x^{2} + 1x + 10x + 2 |
Group the two pairs of terms: |
(5x^{2} + 1x) + (10x + 2) |
Remove common factors from each group: |
x(5x + 1) + 2(5x + 1) |
Notice that the two quantities in parentheses are now identical. That means we can factor out a common factor of (5x + 1): |
(5x + 1)(x + 2) |
Given: |
4x^{2} + 7x 15 |
Find the product ac: |
(4)(-15) = -60 |
Think of two factors of -60 that add up to 7: |
-5 and 12 |
Write the 7x as the sum of -5x and 12x: |
4x^{2} 5x + 12x 15 |
Group the two pairs of terms: |
(4x^{2} 5x) + (12x 15) |
Remove common factors from each group: |
x(4x 5) + 3(4x 5) |
Notice that the two quantities in parentheses are now identical. That means we can factor out a common factor of (4x - 5): |
(4x 5)(x + 3) |
Given a general quadratic trinomial
ax^{2} + bx + c
1. Find the product ac.
2. Find two numbers h and k such that
hk = ac
(h and k are factors of the product of the coefficient of x^{2} and the constant term)
AND
h + k = b
(h and k add to give the coefficient of x)
3. Rewrite the quadratic as
ax^{2} + hx + kx + c
4. Group the two pairs of terms that have common factors and simplify.
(ax^{2} + hx) + (kx + c)
x(ax + h) + (kx + c)
(note: because of the way you chose h and k, you will be able to factor a constant out of the second parentheses, leaving you with two identical expressions in parentheses as in the examples).
· Remember that this wont work for all quadratic trinomials, because not all quadratic trinomials can be factored into products of binomials with integer coefficients. If you have a non-factorable trinomial, you will not be able to do step 2 above.
Why this worksSuppose the quadratic trinomial in question came from multiplying two arbitrary binomials: (px + n)(qx + m) If we multiply this out we will get pqx^{2} + pmx + qnx + nm or pqx^{2} + (pm + qn)x + nm Notice that the coefficient of x consists of a sum of two terms, pm and qn. These are the two numbers we called h and k above. pm = h qn = k Now we see that the two numbers h and k add up to the coefficient of x, which we called b: h + k = b Obviously they are factors of their own product pmqn, but we notice that pq = a, and mn = c, so (pm)(qn) = (pq)(nm) which is equivalent to hk = ac |