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Factoring Worksheet

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In this unit we will study about Factoring. Factoring Worksheets are available on net which will give you a lot of practice questions on factoring. When we talk about any algebraic expression and say that it is the product of two or more smaller algebraic expressions, then each of these smaller expressions is termed as factors. If we write 4* x * y , then 7, x, y are the smaller factors and can not be further factorized. So they are called irreducible factors (and not as prime factors). First method we use for factorizing any algebraic expression is greatest common factor of two or more monomial . by finding the If we have an expression as 12x2y5 and 36x3y3. We first take the numeral part of the two numbers 12 and 36 and take the GCD of these two numbers as 12. Also we observe that x2 and y3 are common values among the two terms. So we get 12 x2y3 as common factor of the two expressions and we get :
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Factoring Worksheet
Factoring Worksheet
In this unit we will study about Factoring. Factoring Worksheets are available on
net which will give you a lot of practice questions on factoring.
When we talk about any algebraic expression and say that it is the product of two
or more smaller algebraic expressions, then each of these smaller expressions is
termed as factors.
If we write 4* x * y , then 7, x, y are the smaller factors and can not be further
factorized. So they are called irreducible factors (and not as prime factors).
First method we use for factorizing any algebraic expression is by finding the
greatest common factor of two or more monomial .
If we have an expression as 12x2y5 and 36x3y3. We first take the numeral part
of the two numbers 12 and 36 and take the GCD of these two numbers as 12.
Also we observe that x2 and y3 are common values among the two terms. So we
get 12 x2y3 as common factor of the two expressions and we get :
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12 x2y3 ( y2 + 3x ) is the solution of factorization of the given expression. This
type of solution is possible only when each term of the expression has a common
factor.
Now we take another situation, when the expression forms the perfect square
and thus we factorize it in the form such that the terms form the perfect square.
For such cases, we must remember the two identities : a2 + b2 + 2 ab = ( a + b )
2 and a2 + b2 - 2 ab = ( a - b )2
Now while solving the expression, we must try to correlate that with which identity
do the expression is related and how will it be expressed in form of perfect
square. Let us take an expression: x2 + 4xy + 4y2
The above expressions can be written as x2 + 2*2xy + (2y)2
Or = x2 + 2*2y * x + (2y)2
This above expression now relates to the standard form of ( a + b )2
Or it can be written as = ( x + 2y )2
More over sometimes the expression is in the form of the difference of two
squares: a2 - b2 = ( a + b) * ( a - b )
Using the above expression solve the 4x2 - 9y2
The above expression can be written as (2x)2 - (3y)2, which satisfies the above
given identity
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So we find that it can be factorized using the formula of ( a + b ) (a - b) and we
get:
(2x)2 - (3y)2 = ( 2x + 3y ) * ( 2x - 3y )
Another type of expressions we find are second degree trinomials, in which we
first arrange the terms of the given expression in the standard form, then we find
the product of the co-efficient of x2 and the constant.
Now the middle term is split in such a form that the two factors so formed gives
the sum equal to the middle term and the product will be equal to the first and the
third term. In x2 + 17x + 60, we can write it as
= x2 + 12x + 5x + 60
= x ( x+ 12 ) + 5 ( x + 12 )
= ( x + 5 ) ( x + 12 ) (Answer).


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