Factoring a Quadratic Trinomial by Grouping
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:
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Given:
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5x2 + 11x + 2
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Find the product ac:
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(5)(2) = 10
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Think of two factors of 10 that add
up to 11:
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1 and 10
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Write the 11x as the sum of 1x
and 10x:
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5x2 + 1x + 10x + 2
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Group the two pairs of terms:
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(5x2 + 1x) + (10x + 2)
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Remove common factors from each
group:
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x(5x + 1) + 2(5x + 1)
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Notice that the two quantities in
parentheses are now identical. That means we can factor out a common factor
of (5x + 1):
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(5x + 1)(x + 2)
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Example:
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Given:
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4x2 + 7x 15
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Find the product ac:
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(4)(-15) = -60
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Think of two factors of -60 that add up to 7:
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-5 and 12
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Write the 7x as the sum of -5x and 12x:
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4x2 5x + 12x 15
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Group the two pairs of terms:
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(4x2 5x) + (12x 15)
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Remove common factors from each
group:
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x(4x 5) + 3(4x 5)
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Notice that the two quantities in
parentheses are now identical. That means we can factor out a common factor
of (4x - 5):
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(4x 5)(x + 3)
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Given a general quadratic trinomial
ax2 + 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 x2 and the constant term)
AND
h + k = b
(h and k add to give the coefficient
of x)
3. Rewrite the quadratic as
ax2 + hx + kx + c
4. 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).
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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.
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Suppose the quadratic trinomial in question came from multiplying two
arbitrary binomials:
(px + n)(qx + m)
If we multiply this out we will get
pqx2 + pmx + qnx + nm
or
pqx2 + (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
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