The topic that we want to discuss is Algebraic Fractions. Students tend to make mistakes while performing operations involving Algebraic Fractions. They could not relate the questions with their prior knowledge in fraction, reciprocal or factorization.
So, let us do some revision on basic concept of fraction first.
If we want to do division involving fraction, we have to change the operation to multiplication and reverse the number that we want to divide.
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Now, what is reciprocal then?
If we have a number
Then, its reciprocal is
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Okay, now we revise on factorization.
1. Factorize ax + bx
In ax and bx, x is common and also a common factor.
Write the common factor x in the polynomial ax + bx outside as x(a + b)
Now, both x and a + b are factors of the polynomial ax + bx
Therefore, factorization of
ax + bx = x(a + b)
2. Factorize a2 + bc + ab + ac
Solution:
let us group terms in a2 + bc + ab + ac
But, which terms to group?
Those terms which yield a common factor on grouping!
Group the terms like this:
a2 + ab + bc + ac
a is the common factor in a2 + ab and b is the common factor in ab + bc
a2 + ab = a (a + b) and bc + ac = c (b + a)
now, the common factor is the binomial (a + b).
Therefore, factorization of
a2 + ab + bc + ac =
a (a + b) + c (a + b) =
(a + b) (a + c)
3. Factorize ax + bx + ay + by
Group terms with similar literal coefficients and numerical coefficients.
One group is ax + bx, in which x is the same literal coefficient,
and the other is ay + by, in which the y is the same literal coefficient,
In ax + bx, the common factor is x and in ay +by, the common factor is y.
on factorization with common factors, we have:
ax + bx = x ( a + b ) and
ay + by = y (a + b)
so ax + bx + ay + by = x (a + b) + y (a + b).
(a + b) is the common binomial factor for the next step of factorization:
(a + b)(x + y)
Therefore, factorization of
ax + bx + ay + by = (a + b)(x + y)
4. Factorize
Students must remember the factorization so that they can answer the questions on algebraic fractions which are usually will involve factorization of two or more terms.
1. Factorize ax + bx
Solution:
In ax and bx, x is common and also a common factor.
Write the common factor x in the polynomial ax + bx outside as x(a + b)
Now, both x and a + b are factors of the polynomial ax + bx
Therefore, factorization of
ax + bx = x(a + b)
2. Factorize a2 + bc + ab + ac
Solution:
let us group terms in a2 + bc + ab + ac
But, which terms to group?
Those terms which yield a common factor on grouping!
Group the terms like this:
a2 + ab + bc + ac
a is the common factor in a2 + ab and b is the common factor in ab + bc
a2 + ab = a (a + b) and bc + ac = c (b + a)
now, the common factor is the binomial (a + b).
Therefore, factorization of
a2 + ab + bc + ac =
a (a + b) + c (a + b) =
(a + b) (a + c)
3. Factorize ax + bx + ay + by
Solution:
Group terms with similar literal coefficients and numerical coefficients.
One group is ax + bx, in which x is the same literal coefficient,
and the other is ay + by, in which the y is the same literal coefficient,
In ax + bx, the common factor is x and in ay +by, the common factor is y.
on factorization with common factors, we have:
ax + bx = x ( a + b ) and
ay + by = y (a + b)
so ax + bx + ay + by = x (a + b) + y (a + b).
(a + b) is the common binomial factor for the next step of factorization:
(a + b)(x + y)
Therefore, factorization of
ax + bx + ay + by = (a + b)(x + y)
4. Factorize
The factorization of difference between two squares will be as below.
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