- Constants and Variable
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- Polynomial expression
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- how to find the degree of a polynomial
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- Value of the polynomial
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- Zeros or roots of the polynomial
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- Adding Polynomials
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- subtracing Polynomials
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- Multiplying Polynomials
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- Dividing Polynomails
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- How to factor polynomials
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- Solved Examples Polynomials
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- polynomial Formative Assignment
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- Dividing Polynomial worksheet
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- Factoring polynomial worksheet
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- Polynomial Test Paper

The various methods to perform factoring of polynomials are

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. $ax^2yz + bxy^2z +cxyz^2$

$=(x \times x \times y \times z \times a) +(x \times y \times y \times z \times b) +(x \times y \times z \times z \times c)$

$=xyz(ax+by+cz)$

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

1. $x^2 + 8x - x - 8$

__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;

$x^2 + 8x = x(x+8)$

What about the last two terms? Observe them. If you take -1 out

$-x-8 = -1(x+8)$

__Step 3__ Putting both part together together,

$x^2 + 8x - x - 8= x(x+8) -1(x+8)$

$= (x+8)(x-1)$

2. $x^5 -2x^3 -2x^2 +4$

__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**

$x^2 + 8x = x(x+8)$

What about the last two terms? Observe them. If you take -1 out

$-x-8 = -1(x+8)$

$x^2 + 8x - x - 8= x(x+8) -1(x+8)$

$= (x+8)(x-1)$

2. $x^5 -2x^3 -2x^2 +4$

$x^5 -2x^3 = x^3(x^2-2)$

What about the last two terms? Observe them. If you take -2 out

$-2x^2 +4 = -2(x^2-2)$

$x^5 -2x^3 -2x^2 +4= x^3(x^2-2) -2(x^2-2)$

$= (x^2-2)(x^3-2)$

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

$P(x) =ax^2 +bx +c$

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**

$4x^2 + 10x - 6$

$=4x^2 +(-2+12)x -6$

$=4x^2 -2x + 12x -6$

$= 2x(2x-1)+ 6(2x-1)$

$= (2x-1)(2x+6)$

- $x^{2} - 5 x - 14 = 0$
- $x^{2} - 3 x - 54 = 0$
- $x^{2} - 14 x + 45 = 0$
- $x^{2} - 8 x - 20 = 0$
- $x^{2} - 11 x + 18 = 0$
- $x^{2} - 13 x + 42 = 0$
- $x^{2} + 3 x - 54 = 0$

First let's find out what is factor theorem

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. $4x^2 + 10x - 6$

$4( x^2 + 5\frac {x}{2} -\frac {3}{2}) = 4(x-a) (x-b)$

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. $P(x)= x^3 +6x^2+11x+6$

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

$x^3 +6x^2+11x+6=K(x+1)(x+2)(x+3)$

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 a

- $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 $

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.$8x^3 + y^3 + 27z^3 -18xyz$

$= (2x)^3 + y^3 + (3z)^3 - 3(2x)(y)(3z)$

$= (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 - (2x)(y) - (y)(3z) - (2x)(3z)]$

$= (2x + y + 3z) (4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6xz)$

$27x^3 +1$

$=(3x+1)(9x^2+1-3x)$

2.$8x^3 + y^3 + 27z^3 -18xyz$

$= (2x)^3 + y^3 + (3z)^3 - 3(2x)(y)(3z)$

$= (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 - (2x)(y) - (y)(3z) - (2x)(3z)]$

$= (2x + y + 3z) (4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6xz)$

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,

$x^2 + y^2 + x^2 -xy -yz -zx \geq 0$

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 $3x + 5y$

$(3x)^2 + (5y)^2 = 181$

$(3x)^2 + (5y)^2 + 30xy - 30xy = 181$

$(3x)^2 + (5y)^2 + 2(5x)(6y) = 181 + 30xy$

$(3x + y)^2 = 181 + 30(-6) = 181 - 180$

$(3x + y)^2 =1$

$3x + y = +1 \; or \; -1$

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

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,

$x^2 + y^2 + x^2 -xy -yz -zx \geq 0$

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 $(a+b+c)^2$

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 $(a+b+c)^2$

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 $x^3 -x^2 +1$

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 $x^51 +x^2 +1$

Check your Answer

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

- 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.
- 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.
- Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
- 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.

Class 9 Maths Class 9 Science

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