Hello friends, today we are going to learn about the Factoring Expressions.
Factoring is just the reverse of expanding but a bit complex than a (b + c) = ab + ac, but the procedure followed is the same.
Factoring is generally an expression which is written as the product of two or more expressions.
For Example - 15 = 5 x 3, where 5 and 3 are both the factors of 15. The most basic ways to factor an expression is by finding the common factors.
Now what are common factors?
If each term of the expression has several factors and has atleast one factor that is common in them, then the common term is known is its common factor.
For Example – x2 y4 + 2 x2.
In this expression we can see that x2 is common in both the terms. So it can be written as
x2 ( y4 + 2 ). So now x2 is the common factor of the above expression or it can also be said as it is the Greatest Common Factor( GCF) of the above expression.
The general way of writing the common factors is GCF x ( remaining expression).
We take another example for more understanding.
Let the expression be 8Y3B2 + 16Y2B.
Now to find the common factors of this we have to see that what is common, so it seems like the term 8Y2B is common in the above expression. So it can be written as 8Y2B ( YB + 2 ). This is the same way as we have written the general formula GCF x ( remaining expression).
Now as I said above factoring is just opposite of expanding. So let's now verify this.
We take two terms ( x + 5) ( x + 1).
The expansion of this is x2 + 6x + 5. And the factoring of this answer gives us ( x + 5) ( x + 1). So these are the factors of the given expression. We do this by splitting the x term in two parts. There are different ways of expanding but one among them helps us to find the common factor.
The 6x can be broken as 5x+1x, 2x+4x and 3x +3x. It satisfies x + 5x, so that we get our common factors. For doing this we have to follow some general rules.
Let the general quadratic equation be ax2 + bx + c. So the general rules are.
1.multiply the constants and c without forgetting their signs.
2.Write down b, with its sign.
3.Write all the possible factors of the product of ac and find which pairs add up to give b .
4.now rewrite the x terms as the sum of two terms with these selected numbers as the coefficients
So by this we can find the factors of the quadratic expression. The above method we followed is known as the decomposition method.
To make perfect squares. Any expression of the form x2 + 2ax + a2 can be written as a perfect square (x + a)2. To see whether an expression is a perfect square or not we first see that whether the constant term in it is a whole square or not, if it is a perfect square then take square root of it and multiply it by 2. If the result is equal to the coefficient of x then the term could be easily written as a perfect square.
For Example – Factorize x2 + 8x + 16.
We can see that the constant term is 16, and it is a perfect square of 4.
Now 2 x 4 is equal to 8 which is the coefficient of x. So our given expression can be written as
(x + 4)2.
Now we move to differences of square.
The differences of the square term are written in a general way as (something)2 – (something)2.
Let's understand it with the help of a simple example. Let's take (a2x2 – b2), it can be factorised as (ax + b)(ax – b). We can see that the factors are identical except the sign between them.
For Example-
Factorize 9 – r2.
To factorize it we use the above expression as the general function and find the value of a, x and b.
Here a2 = 1, thus a = 1
x2 = 9, thus x = 3
b2 = r2, thus b = r,
so the above expression can be written as ( 3 + r ) ( 3 – r ). To understand it more we will take another example.
Factorize 9p2 – 4.
Here also we follow the above method and find the values of a, x and b.
a2 = 9, thus a = 3x2 = p2, thus x = pb2 = 4, thus b = 2,
so the above expression after factorization gives ( 3p + 2 ) ( 3p – 2 ).
There is one more thing that every expression cannot be factorized by this method, so we use the Quadratic Formula to find the roots and then we write the factors.
Now we move to the Difference of cubes.
It is very much similar to the difference of squares, the only difference is that it has cubes in the exponents in place of squares.
We write it in a general way as a3 – b3. Which can be factorized as (a - b)(a2 - ab + b2). The factorization seems to be very difficult to remember, so what we have to remember is that if we see an expression like a3 – b3, then (a – b) is a common factor and the rest we can find out using the long division method.
Sum of Cubes.
The general way of writing this is a3 + b3. And when factorized we see that the result is
(a + b)(a2 - ab + b2). In this also we have to remember only that when we see an expression like
a3 + b3, then we should remember that (a + b) is its common factor and the rest we can find out using the long division method.
Now as we have performed factorization on expressions with square and cube exponents, so now we move to the expressions with higher degrees. Sometimes students get panicked seeing higher degrees, but there is no need to panic as the expressions with higher degrees are as simple as the quadratic and cubic ones. There will always be a common factor in them because of which the remaining term will be converted into a quadratic expression.
For Example take (9x4 - 2x3 + 10x2) as your expression.
Now we can see that x2 is common from the above expression. So we can finally write it as
x2(9x2 - 2x + 10). Now the expression within the brackets can be solved by using the decomposition method. If we see that there is a constant term in it, then the simplest way is by using the long division method.
Let's understand this with the help of an example.
Solve x3 - 2x2 - 5x + 6 with (x + 2) as one of its factors.
We will use the long division method to solve the above expression.
x2 - 4x + 3
-------------------
(x + 2)| x3 - 2x2 - 5x + 6)
x3 + 2x2
---------
-4x2 - 5x
-4x2 - 8x
---------
3x + 6
3x + 6
------
0
so the factor other than x+2 is x2 - 4x + 3.
now we have to solve x2 - 4x + 3 by using the decomposition method.
So now split -4x as -x and -3x. So we get the factors as (x - 1) and (x – 3).
so at the end the answer to x3 - 2x2 - 5x + 6 is (x + 2)(x - 1)(x – 3).
In this question we were given one factor of the expression, what if there is no factor given in such type of questions. Then we could still factorize by assuming that ( x – k) as its common factor. Where k is some value which satisfies the given equation. If we find the value of k by hit and trial then we wil get our one factor and the rest could be done by long division method.
So let's understand this with the help of an example.
Solve x3 + 1.
So we have no common expression given with this question.
We assume the one factor as (x – k), now by hit and trial we can see that -1 satisfies the above equation so our factor is (x + 1). Now by using long division method we can find the rest.
x2 - x + 1
-----------------
(x + 1)| x3 + 0x2 + 0x + 1)
x3 + x2
-------
-x2 + 0x
-x2 - x
-------
x + 1
x + 1
-----
0
So the another factor is x2 - x + 1.
Now the solution to x3 + 1 is equal to ( x + 1) (x2 - x + 1).
So now I hope that you would be able to perform the operations on Factoring expressions
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