On this page "factoring polynomials",we are going see clear explanation of about this topic with example problems.

Grouping means factor out the common terms from the group.Let us see some example problems to understand the topic factoring polynomials with two variables.

This can be done by the following methods

(ii) Factor using algebraic identities

(iii) Factoring quadratic equations

**Example 1:**

Factor pq - pr - 3ps

**Solution:**

= pq - pr - 3ps

To factor the above algebraic expression,first we have ask a question for ourself

Question :

Do we find any common term in the above algebraic expression ?

Answer:

Yes, we have " p" in all three terms.

So the answer is p (q - r- 3s)

Now we are going to see the next example of the topic"factoring polynomials".

**Example 2:**

Factor 4a - 8b + 5ax - 10bx

**Solution:**

= 4a - 8b + 5ax - 10bx

We have 4 terms in the given algebraic expression.Now we are going to split into two groups.

4a and -8b are in one group (Group 1)

5ax and -10bx are in the other group(Group 2)

Question :

Do we have any common variable in group 1?

Answer: No

Question :

Do we have any common variable in group 2?

Answer: Yes,that is x.

Question :

Do we have any common number in group 1?

Answer:

Yes, we can split 8 as multiple of 4.

Question :

Do we have any common number group 2?

Answer:

Yes, we can split 10 as multiple of 5.

Factor out the common term

So, the factors are (a - 2b) (4 + 5x)

Now we are going to see the next example of the topic"factoring polynomials".

**Example 3:**

Factor 2a**³ - 3a****²b + 2****a****²c **

**Solution :**

= 2a³ - 3a²b + 2a²c

Question :

Do we find any common variable or number in all three terms?

Answer: yes,we have a² in all three terms.

= a² (2a - 3b + 2c)

Now we are going to see the next example of the topic"factoring polynomials".

**Example 4:**

Factor 10x³ - 25x⁴ y** **

**Solution :**

= 10x³ - 25 x⁴ y** **

Question :

Do we find any common variable or number in all three terms?

Answer: yes,we have x³ and for number we can split both 10 and 25 as the multiple of 5.So we have 5x³ as common term.

= 5x³ (2 - 5xy)

Now we are going to see the next example of the topic"factoring polynomials".

**Example 5:**

Factorize 9x² - 24xy + 16y²

**Solution:**

**We have x****² as the first term and y****² as the last term.Since there are only three terms.We can compare the given question with the algebraic identity a****² - 2ab + b****²**

** = 9x² - 24xy + 16y²**

= 3**² **x**² - 2(3x) (4y) + 4****² ****y****²**

** = (3 x)² - 2(3x) (4y) + (4 y)²**

** = (3x - 4y)**

**Now we are going to see the next example of the topic"factoring polynomials".**

**Example 6:**

Factorize 64 a³ - 343 b³

**Solution:**

**We can split 64 as 4 x 4 x 4 and we can split 343 as 7 x 7 x 7.**

** a³ - b³ = (a - b) (a² + ab + b²)**

by comparing the given polynomial (4a)³ - (7b)³,we get

= (4a - 7b) [(4a)² + (4a) (7b) + (7b)²]

= (4a - 7b) (16a² + 28ab + 49b²)

Now we are going to see the next example of the topic"factoring polynomials".

Factoring quadratics with a leading coefficient of 1

In a quadratic "Leading coefficient" means "coefficient of x²".

(i) If the coefficient is 1 we have to take the constant termand we have to split it as two parts.

(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.

(iii) Now we have to write these numbers in the form of (x + a) and (x +b)

Factoring quadratic equations when a isn't 1

(i) If it is not 1 then we have to multiply the coefficient of x² by the constant term and we have to split it as two parts.

(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.

(iii) Divide the factors by the coefficient of x². Simplify the factors by the coefficient of x² as much as possible.

(iv) Write the remaining number along with x.

Let us see some examples for better understanding.

**Example 7:**

Factor x² + 17 x + 60

**Solution:**

**In the first step we are going to check whether we have 1 as the coefficient of x² or not. **

**Since it is 1. We are going to take the last number. That is 60 and we are going to factors of 60.**

**All terms are having positive sign. So we have to put positive sign for both factors.**

Here,

10 x 6 = 60 but 10 + 6 = 16 not 17

15 x 4 = 60 but 15 + 4 = 19 not 17

12 x 5 = 60 and 12 + 5 = 17

2 x 30 = 60 but 2 + 30 = 32 not 17

**(x + 12) (x + 5)** are the factors of x² + 17 x + 60.

More examples of factoring quadratic equation

**Example 8:**

Factor 2 x² + x - 6

**Solution :**

To factor this quadratic equation we have to multiply the coefficient of x² by the constant term

So that we get -12, now we have to split -12 as the multiple of two numbers.

Since the last term is having negative sign.So we have to put negative sign for the least number.

Now we have to divide the two numbers 4 and -3 by the coefficient of x² that is 2.If it is possible we can simplify otherwise we have to write the numbers along with x.

So (x + 2) (2x - 3) are the factors of 2 x² + x - 6

**More examples of factoring quadratic equation when the leading coefficient of x² is not 1.**

Related topics

- Factoring quadratic equations when the coefficient of x² is not 1
- Factoring using Algebraic identities
- Solving quadratic equation using sompleting the square method
- Solving quadratic equation by using quadratic formula
- Practical problem using quadratic equations

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