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.
|Given:||5x2 + 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:||5x2 + 1x + 10x + 2|
|Group the two pairs of terms:||(5x2 + 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:||4x2 + 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:||4x2 – 5x + 12x – 15|
|Group the two pairs of terms:||(4x2 – 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 ax2 + bx + c
- Find the productac.
- Find two numbershand k such that
hk = ac (h and k are factors of the product of the coefficient of x2 and the constant term)
h + k = b (h and k add to give the coefficient of x)
- Rewrite the quadratic asax2+ hx + kx + c
- Group the two pairs of terms that have common factors and simplify.
(ax2+ 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 won’t 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 works
Suppose the quadratic trinomial in question came from multiplying two arbitrary binomials:
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
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