# How to factor polynomials

Table of Content

## How to factor polynomials

We have studied about polynomial addition,subtraction ,multiplication and division. Now it time to learn How to factor polynomials. As you are already aware factoring is the process by which we go about determining what we multiplied to get the given quantity. So factoring polynomials means finding the terms which when multiplied together produced the polynomials
The various methods to perform factoring of polynomials are

### Greatest Common Factor

We can look at each of the term in the polynomials ,factorize each term and then find common factors to factorize the expression. This should be the first steps always when you are factoring the polynomials

In these problem, we need to factorize each term and then find common factors to factorize the expression
i. $7x-42$
$=(7 \times x) - (7 \times 6)$
$=7(x-6)$

ii. $6p-12$
$=(6 \times p) - (6 \times 2)$
$=6(p-2)$

iii. $7a^2+14a$
$=(7 \times a \times a) + (2 \times 7 \times a)$
$=7a(a+2)$

iv. $-16z+ 20z^3$
$=4z(5z^2-4)$

v. $20 l^2 m + 30 a l m$
$=(2 \times 2 \times 5 \times l \times m \times l) + (2 \times 3 \times 5 \times a \times l \times m)$
$=2lm(10l+15a)$

vi. $5 x^2y -15 xy^2$
$=(5 \times x \times x \times y) + (5 \times 3 \times x \times y \times y)$
$=5x(xy-3y^2)$

vii. $10a^2 -15 b^2+20c^2$
$=(2 \times 5 \times a \times a) - (5 \times 3 \times b \times b) + (5 \times 2 \times 2 \times c \times c)$
$=5(2a^2-3b^2+4c^2)$

viii. $-4a^2 +4a b-4ca$
$=(-4 \times a \times a) +(4 \times b \times a) - (4 \times c \times a)$
$=-4a(a-b+c)$

ix. $x^2yz + xy^2z +xyz^2$
$=(x \times x \times y \times z) +(x \times y \times y \times z) +(x \times y \times z \times z)$
$=xyz(x+y+z)$
x. $ax^2yz + bxy^2z +cxyz^2$
$=(x \times x \times y \times z \times a) +(x \times y \times y \times z \times b) +(x \times y \times z \times z \times c)$
$=xyz(ax+by+cz)$

Practice Questions
• $24p^5-20p^3-10p^2$
• $2x^4 -4x^2 -8$
• $x^yz+xy^2z+xyz^2$
• $81-9x^2$
• $11z^3 - 121z+55$

### factor polynomials by grouping

When we don't see common factor across all the terms, we may look at grouping the terms and check if we find binomial factor from both the groups .Let us take example to take close look at the method
1. $x^5 -2x^3 -2x^2 +4$
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor x;
$x^5 -2x^3 = x^3(x^2-2)$
What about the last two terms? Observe them. If you take -2 out
$-2x^2 +4 = -2(x^2-2)$
Step 3 Putting both part together together,
$x^5 -2x^3 -2x^2 +4= x^3(x^2-2) -2(x^2-2)$
$= (x^2-2)(x^3-2)$

Practice Questions
• $z^{6} + 6 z^{5} + 4 z + 24$
• $y^{3} + 8 y^{2} + 9 y + 72$
• $z^{5} + 4 z^{3} + z^{2} + 4$
• $x^{3} + 7 x^{2} + 9 x + 63$
• $x^{6} + 4 x^{5} + 2 x + 8$
• $z^{5} + 2 z^{4} + 4 z + 8$

### factor Quadratic polynomials by splitting the middle term

$P(x) =ax^2 +bx +c$
Let its factors be $(px + q)$ and $(rx + s)$. Then
$ax^2 +bx +c= (px + q) (rx + s) = pr x^2 + (ps + qr) x + qs$
Comparing the coefficients of $x^2$, we get $a = pr$.
Similarly, comparing the coefficients of x, we get $b = ps + qr$.
And, on comparing the constant terms, we get $c = qs$
This shows us that b is the sum of two numbers ps and qr, whose product is
$(ps)(qr) = (pr)(qs) = ac$
Therefore, to factorise $ax^2 +bx +c$, we have to write b as the sum of two numbers whose product is ac
1. $4x^2 + 10x - 6$
Step 1 Here b=10 and ac=-24
Step 2 We need to find factor of -24,whose sum is 10, the number which satisfy this is 12 and -2
$4x^2 + 10x - 6$
$=4x^2 +(-2+12)x -6$
$=4x^2 -2x + 12x -6$
$= 2x(2x-1)+ 6(2x-1)$
$= (2x-1)(2x+6)$

Practice Questions
1. $x^{2} - 5 x - 14 = 0$
2. $x^{2} - 3 x - 54 = 0$
3. $x^{2} - 14 x + 45 = 0$
4. $x^{2} - 8 x - 20 = 0$
5. $x^{2} - 11 x + 18 = 0$
6. $x^{2} - 13 x + 42 = 0$
7. $x^{2} + 3 x - 54 = 0$

### 4) Factorising a Polynomial by Factor Theorem

First let's find out what is factor theorem

#### Factor's Theorem's

If x-a is a factor of polynomial p(x) then p(a)=0 or if p(a) =0,x-a is the factor the polynomial p(x)
We can use this theorem to find factors of the polynomials

Lets first take a look at quadratic polynomial
1. $4x^2 + 10x - 6$
Step 1 Take 4 out of the expression $4( x^2 + 5\frac {x}{2} -\frac {3}{2})$
Step 2 This can be further expressed as
$4( x^2 + 5\frac {x}{2} -\frac {3}{2}) = 4(x-a) (x-b)$
Here so ab= -3/2
We need to look for all the factor of the term -3/2 i.e +1/2,-1/2,-3,+3,1/4,-1/4,3/2,-3/2
Wherever it satisfies the factor theorem, we are good
In this particular case
P(1/2)=P(-3)=0, we can write like this

$4( x^2 + 5\frac {x}{2} -\frac {3}{2}) = 4(x- \frac {1}{2}) (x+3)$
$= (2x-1)(2x+6)$
For the last example, it may seems splitting the terms seems more efficient
Now let us take a look at cubic polynomial
Suppose the Polynomial is the form
2. $P(x)= x^3 +6x^2+11x+6$

Step 1
We need to look at the constant 6 and factorise it
The factor of 6 will be 1,2,3
Now we can try the polynomial for all the values -3,-2,-1,1,2,3
Wherever it satisfies the factor theorem, we are good
In this particular case
$P(-1)=P(-2)=P(-3)=0$, we can write like this
Step 2
$x^3 +6x^2+11x+6=K(x+1)(x+2)(x+3)$

We can put any value of x in this identity and get the value of x
In this particular case putting x=0, we get K=1

So the final identity becomes
$x^3 +6x^2+11x+6=(x+1)(x+2)(x+3)$

Or we can just find the one factor (x+1) and try the splitting or long division method to get the remaining quadratic polynomial and solve that by splitting method
$x^3 +6x^2+11x+6$
$= x^3 +x^2+5x^2+5x+6x +6$
Splitting cubic polynomial to get (x+1)
$=x^2(x+1) + 5x(x+1) + 6(x+1)$
$=(x+1)(x^2+5x+6)$
Now solving the quadratic polynomial by splitting method
$=(x+1)(x^2+2x+3x+6)$
$=(x+1)[x(x+2)+3(x+2)]$
$=(x+1)(x+2)(x+3)$
In General Term,

$S(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+....+ax+a_0$

Look for the factors in a0/an, Take both the positive and negative values and find out which suites your polynomial and then find the value of k

Practice Questions
• $x^{3} + 22 x^{2} + 140 x + 200$
• $x^{3} + 4 x^{2} + 5 x + 2$
• $x^{3} + 13 x^{2} + 48 x + 36$
• $x^{3} + 15 x^{2} + 72 x + 108$
• $x^{3} + 4 x^{2} + 5 x + 2$
• $x^{3} + 7 x^{2} + 15 x + 9$
• $x^{3} + 14 x^{2} + 60 x + 72$
• $x^{3} + 12 x^{2} + 39 x + 28$
• $x^{3} + 7 x^{2} + 11 x + 5$
• $x^{3} + 17 x^{2} + 88 x + 144$
• $x^{3} + 14 x^{2} + 43 x + 30$
• $x^{3} + 9 x^{2} + 26 x + 24$

### Factoring polynomial using Algebraic Identities

Identity I : $(x + y)^2 = x^2 + 2xy + y^2$
Identity II : $(x - y)^2 = x^2 - 2xy + y^2$
Identity III : $x^2 - y^2 = (x + y) (x - y)$
Identity IV : $(x + a) (x + b) = x^2 + (a + b)x + ab$
Identity V :$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
Identity VI : $(x + y)^3 = x^3 + y^3 + 3xy (x + y)$
Identity VII :$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$
Identity VIII : $x^3 + y^3 + z^3 -3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
Identity IX : $x^3 + y^3 = (x + y )(x^2 + y^2 - xy )$
Identity X : $x^3 - y^3 = (x - y )(x^2 + y^2 + xy )$
1. $27x^3 +1$
Step 1 Both th terms are exact cubes (3x) and 1
Step 2 We can the algebraic identity IX
$27x^3 +1$
$=(3x+1)(9x^2+1-3x)$

2.$8x^3 + y^3 + 27z^3 -18xyz$
$= (2x)^3 + y^3 + (3z)^3 - 3(2x)(y)(3z)$
$= (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 - (2x)(y) - (y)(3z) - (2x)(3z)]$
$= (2x + y + 3z) (4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6xz)$

## Solved Examples Polynomials

1. If $x + \frac {1}{x} = 12$, evaluate $x^2 + \frac {1}{x^2}$

Solution $x + \frac {1}{x} = 12$
Squaring both the sides
$(x + \frac {1}{x})^2 = (x)^2 +\frac {1}{x^2} + 2(x)(1/x) = 144$
$x^2 + \frac {1}{x^2} + 2 = 144$
$x^2+ \frac {1}{x^2} = 144 - 2 = 142$

2. If $9x^2+ 25y^2 = 181$ and $xy = -6$. Find the value of $3x + 5y$

Solution $9x^2+ 25y^2 = 181$
$(3x)^2 + (5y)^2 = 181$
$(3x)^2 + (5y)^2 + 30xy - 30xy = 181$
$(3x)^2 + (5y)^2 + 2(5x)(6y) = 181 + 30xy$
$(3x + y)^2 = 181 + 30(-6) = 181 - 180$
$(3x + y)^2 =1$
$3x + y = +1 \; or \; -1$

3. Prove that $x^2 + y^2 + x^2 -xy -yz -zx$ is always non-negative for all values of x, y and z.

Solution To prove $x^2 + y^2 + x^2 -xy -yz -zx \geq 0$
We know that square of any number (+ve or -ve) is always +ve.
$x^2 + y^2 + z^2 -xy -yz -zx$
$= \frac {1}{2}[2x^2 + 2y^2 + 2z^2 -2xy -2yz -2zx]$
$=\frac {1}{2}[x^2 + y^2 -2xy + y^2 + z^2 -2yz + x^2 + z^2 -2zx]$
$=\frac {1}{2}[(x^2 + y^2 -2ab) + (y^2 + z^2 -2yz) + (x^2 + z^2 -2zx)]$
$=\frac {1}{2}[(x - y)^2 + (y - z)^2 + (z - x)^2 ]$
Here, all the terms are always be positive,
$x^2 + y^2 + x^2 -xy -yz -zx \geq 0$

## Practice Questions

1. If ( x - 4) is a factor of the polynomial $2x^2 + Ax + 12$ and ( x - 5) is a factor of the polynomial $x^3 - 7x^2+ 11 x + B$ , then what is the value of ( A - 2B )?
2. if $x - \frac {1}{x} =3$ then find the value of $x^3 -\frac {1}{x^3}$
3. if $a+b+c=7$ and $ab+bc+ca=20$ find the value of $(a+b+c)^2$
Cross-word Puzzle to test your knowledge on polynomials

Across
3. A polynomial of two terms is called
5. A polynomial of degree three is called a ____ polynomial.
8. What is called the expression $3x^2+x -1$ in $3x^2 + x - 1 = (x + 1) (3x - 2) + 1$
9. A polynomial of three terms is called
10. what is called -1 for $x^2$ in the expression $x^3 -x^2 +1$
Down
1. A polynomial of one term is called
2. what is the last term (1) in the expression $3x^2 + x - 1 = (x + 1) (3x - 2) + 1$
4. A polynomial of degree two is called a _____ polynomial.
6. A polynomial of degree one is called a ____ polynomial.
7. What is called 51 in expression $x^{51} +x^2 +1$

• Notes
• NCERT Solutions
• Assignments
• Revision sheet

Reference Books for class 9 Math

Given below are the links of some of the reference books for class 9 Math.

1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

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