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Factors are the pair of natural numbers which give the resultant number.
Example
24 = 12 × 2 = 6 × 4 = 8 × 3 = 2 × 2 × 2 × 3 = 24 × 1
Hence, 1, 2, 3, 4, 6, 8, 12 and 24 are the factors of 24.
If we write the factors of a number in such a way that all the factors are prime numbers then it is said to be a prime factor form.
The prime factor form of 24 is
24 = 2 × 2 × 2 × 3
Like any natural number, an algebraic expression is also the product of its factors. In the case of algebraic expression, it is said to be an irreducible form instead of prime factor form.
7pq = 7 × p × q =7p × q = 7q × p = 7 × pq
These are the factors of 7pq but the irreducible form of it is
7pq = 7 × p × q
2x (5 + x)
Here the irreducible factors are
2x (5 + x) = 2 × x × (5 + x)
The factors of an algebraic expression could be anything like numbers, variables and expressions.
As we have seen above that the factors of algebraic expression can be seen easily but in some case like 2y + 4, x^{2 }+ 5x etc. the factors are not visible, so we need to decompose the expression to find its factors.
1. Method of Common Factors
In this method, we have to write the irreducible factors of all the terms
Then find the common factors amongst all the irreducible factors.
The required factor form is the product of the common term we had chosen and the leftover terms.
Sometimes it happens that there is no common term in the expressions then
We have to make the groups of the terms.
Then choose the common factor among these groups.
Find the common binomial factor and it will give the required factors.
Factorise 3x^{2 }+ 2x + 12x + 8 by regrouping the terms.
Solution:
First, we have to make the groups then find the common factor from both the groups.
Now the common binomial factor i.e. (3x + 2) has to be taken out to get the two factors of the expression.
Remember some identities to factorise the expression
(a + b)^{2} = a^{2} + 2ab + b^{2}
(a – b)^{2} = a^{2} – 2ab + b^{2}
(a + b) (a – b) = a^{2} – b^{2}
We can see the different identities from the same expression.
(2x + 3)^{2} = (2x)^{2}+ 2(2x) (3) + (3)^{2}
= 4x^{2} + 12x + 9
(2x - 3)^{2} = (2x)^{2} – 2(2x)(3) + (3)^{2}
= 4x^{2 }-12x + 9
(2x + 3) (2x - 3) = (2x)^{2} – (3)^{2}
= 4x^{2} - 9
Example 1
Factorise x– (2x – 1)^{2} using identity.
This is using the identity (a + b) (a – b) = a^{2} – b^{2}
x^{2}- (2x - 1)^{2} = [(x + (2x - 1))] [x – (2x-1))]
= (x + 2x - 1) (x – 2x + 1)
= (3x – 1) (- x + 1)
Example 2
Factorize 9x² - 24xy + 16y² using identity.
1. First, write the first and last terms as squares.
9x² - 24xy + 16y²
= (3x)^{2} – 24xy + (4y)^{2}
2. Now split the middle term.
= (3x)^{2} – 2(3x) (4y) + (4y)^{2}
3. Now check it with the identities
4. This is (3x – 4y)^{2}
5. Hence the factors are (3x – 4y) (3x – 4y).
Example 3
Factorise x^{2} + 10x + 25 using identity.
x^{2} + 10x + 25
= (x)^{2} + 2(5) (x) + (5)^{2}
We will use the identity (a + b)^{ 2} = a^{2} + 2ab + b^{2} here. Therefore,
x^{2} + 10x + 25 = (x + 5)^{2}
(x + a) (x + b) = x^{2} + (a + b) x + ab.
Example:
Factorise x^{2} + 3x + 2.
If we compare it with the identity (x + a) (x + b) = x^{2} + (a + b) x +ab
We get to know that (a + b) = 3 and ab = 2.
This is possible when a = 1 and b = 2.
Substitute these values into the identity,
x^{2} + (1 + 2) x + 1 × 2
(x + 1) (x + 2)
Division is the inverse operation of multiplication.
Write the irreducible factors of both the monomials
Cancel out the common factors.
The balance is the answer to the division.
Solve 54y^{3} ÷ 9y
Write the irreducible factors of the monomials
54y^{3} = 3 × 3 × 3 × 2 × y × y × y
9y = 3 × 3 × y
Write the irreducible form of the polynomial and monomial both.
Take out the common factor from the polynomial.
Cancel out the common factor if possible.
The balance will be the required answer.
Solve 4x^{3} + 2x^{2} + 2x ÷ 2x.
Write the irreducible form of all the terms of polynomial
4x^{3} + 2x^{2} + 2x
= 4(x) (x) (x) + 2(x) (x) + 2x
Take out the common factor i.e.2x
= 2x (2x^{2} + x + 1)
In the case of polynomials we need to reduce them and find their factors by using identities or by finding common terms or any other form of factorization. Then cancel out the common factors and the remainder will be the required answer.
Solve z (5z^{2} – 80) ÷ 5z (z + 4)
Find the factors of the polynomial
= z (5z^{2} – 80)
= z [(5 × z^{2}) – (5 × 16)]
= z × 5 × (z^{2} – 16)
= 5z × (z + 4) (z – 4) [using the identity a^{2 }– b^{2} = (a + b) (a – b)]
Some Common Errors
While adding the terms with same variable students left the term with no coefficient but the variable with no coefficient means 1.
2x + x + 3 = 3x + 3 not 2x +3
We will consider x as 1x while adding the like terms.
If we multiply the expressions enclosed in the bracket then remember to multiply all the terms.
2(3y + 9) = 6y + 18 not 6y + 9
We have to multiply both the terms with the constant.
If we are substituting any negative value for the variables then remember to use the brackets otherwise it will change the operation and the answer too.
If x = – 5
Then 2x = 2(-5) = -10
Not, 2 – 5 = - 3
While squaring of the monomial we have to square both the number and the variable.
(4x)^{2} = 16x^{2} not 4x^{2}
We have to square both the numerical coefficient and the variable.
While squaring a binomial always use the correct formulas.
(2x + 3)^{2 }≠ 4x^{2}+ 9 But (2x + 3)^{2} = 4x^{2 }+ 12x + 9
While dividing a polynomial by a monomial remember to divide each term of the polynomial in the numerator by the monomial in the denominator.
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