FACTORISATION STEP BY STEP
Factoring by Grouping
We can sometimes factor a polynomial by making creative use of
the Distributive Property
Example 1:
Factor 2xy – 6xz + 3y – 9z.
You can get a clue from the coefficients: we have a 2 and a –6, and we also
have a 3 and a –9. There is a proportional relationship here .
Setp 1 : Take the common Factor 2x out of the first two terms:
2xy – 6xz + 3y – 9z = 2x(y – 3z) + 3y – 9z
Step 2: Then take the common factor 3 out of the second two terms.
2xy – 6xz + 3y – 9z
= 2x(y – 3z) + 3(y – 3z)
Step 3 :Since the same quantity y – 3z appears twice, we can use the
distributive property to write this more simply:
2xy – 6xz + 3y – 9z = (2x + 3)(y – 3z)
Example 2:
Factor x2 + xy + 3x + 3y.
Step 1: Group the terms as follows:
x2 + xy +
3x + 3y = (x2 + 3x) + (xy + 3y)
=
x (x+3) + y( x+3)
= (x + y)(x + 3)
Example 3
Factor xy − 4y − 3x + 12
Solution:
Group the first two and the last two terms together, and factor
out the common factor from each grouping:
(xy − 4y) + (−3x + 12) = y(x − 4) + 3(−x + 4)
This time there is no common factor. Try again, this time
factoring a −3 from the last grouping. This works!
xy − 4y − 3x + 12 = y(x − 4) − 3(x − 4)
= (x − 4)(y − 3)
EXERCISES.
1.
Factor each of the following by grouping:
xy + 7x + 4y + 28
= x(
) + 4(
=(
)(
2.
2xy + 5x + 10y + 25
)
)
3.
x3 + 3x2 + 9x + 27
4.
x3 − 3x2 + 9x – 27
5.
ax + bx + ay + by
6.
ax + ac + bx + bc
7.
ax − bx + ay − by
8.
ax − ac + bx − bc
9.
ax − bx − ay + by
10.
= x (a − b) − y ( a − b)
=(
11.
)(
)
ax − ac − bx + bc
= ___(
=(
xy − 5x − 2y + 10
12.
= xy(
=
=
x3 − x2 − 9x + 9
= x2 (
) − 9(
=(
)(
=(
)(
14.
)
)
)(
)(
)
) − 5(
)
x3 − 5x2 − 4x + 20
= x2 (
=
)
)
x2y + xy2 − 5x − 5y
=
13.
) − ___(
=
) − 4(
)
15.
x3 + 7x2 − x − 7
= x2 (
) − 1(
16.
x3 − 5x2 + 25x − 125
)
17.
x3 + 5x2 − 25x − 125
18.
x3 − 5x2 − 25x + 125
19.
x3 − 4x2 + 9x − 36
20.
x3 + 4x2 − 9x − 36
Factoring x2 + bx + c
Factor n2 + 8n + 15.
This can be done (factored) by finding two numbers whose sum is 8 and
product is 15: use 3 and 5. So we get:
n2 + 8n + 15
= n2 + 3n + 5n + 15 . . . . now take common factor out in pairs
= n(n + 3) + 5(n + 3) . . . .use the Distributive Property
= (n + 5)(n + 3)
Factor x2 + 6x – 16.
Here we need to find two numbers with opposite signs which have –16 as a
product and 6 as a sum.
The factor pairs for –16 are:
–16 = (–16)(1) ; –16 + 1 = –15
–16 = (–8)(2) ; –8 + 2 = –6
–16 = (–4)(4) ; –4 + 4 = 0
–16 = (–2)(8) ; –2 + 8 = 6
–16 = (–1)(16); –1 + 16 = 15
–2 and 8 work. So we can factor the polynomial as
x2 + 6x – 16 = (x – 2)(x + 8).
Example:
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)
Example:
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)
EXERCISES. Factor completely, using the sum and difference of cubes
formulas.
1.
x3 − 8
2.
x3 − 125
= x3 − 23 [x=first; 2=second]
=(
−
=(
=(
)(x2 + 2x + 22)
)(
= ( − )(
)
=(
3. x3 − 64
4
=(
)3 − (
=(
)(
)3
)
5. x3 + 8
=(
.
+
)(x2 −
+
)(
)3 − (
=(
)(
6.
x3 + 64
=(
)3 + (
=(
)(
+
=(
x3 + 125
Q
)3
)
)
)
=(
=(
)
+
)(
=(
)
+
)3
x3 − 27
= x3 + 23 [x=first; 2=second]
=(
)3 − (
)3
)
)3 + (
)(
−
)(
)3
+
)
)
x3 + 27
=(
)3 + (
=(
)(
)3
)
Q
8x3 − 125
= (2x)3 − 53
=(
−
27x3 − 8y3
[2x=first; 5=second]
=(
)[(2x)2 + (2x)(5) + 52]
=(
Q
Q
)(
=(
)
64x3 + 125
=(
Q
)3 − (
−
=(
)[
]
=(
)[
=(
)(
)
=(
)(
39.
8x3 + 1
+
+
)
27x3 + 8y3
)3 + (
8x3 − 27y3
)[
)(
=(
Q.
)3
)3
=(
Q.
)3 + (
)3
]
)
125y3 − 8x3
40.
125y 3 −
1
In the next exercises, don't forget to factor the common factor first.
Q.
16x4 − 54x
Q.
3x3 − 24y3
= 2x(8x3 − 27)
=
= 2x[( 2x )3 − ( 3 )3 ]
= 2x(___ − ___)(____+____+____)
Q.
Q.
5x4 + 40x
3x5y5 − 81x2y2
=
=
Q.
Q
10x5y + 80x2y4
16x2y2 + 250x2y5
]