Welcome to Class 9 Maths chapter 2 Polynomials notes.The topics in this page are How to factor polynomials using Greatest Common Factor,Grouping ,splitting middle term ,Factor theorem And Algebraic Identities along with examples, and questions and answers .This is according to CBSE and the NCERT textbook. If you like the study material, feel free to share the link as much as possible.
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
(1)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
$24p^5-20p^3-10p^2$
$2x^4 -4x^2 -8$
$x^yz+xy^2z+xyz^2$
$81-9x^2$
$11z^3 - 121z+55$
(2)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.This will be specially useful on when you are trying to factorize polynomials with 4 or more terms.Let us take example to take close look at the method
how to factor polynomials with 4 terms
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,
$x^5 -2x^3 -2x^2 +4= x^3(x^2-2) -2(x^2-2)$
$= (x^2-2)(x^3-2)$
(3)Factor Quadratic polynomials by splitting the middle term
Let the Quadratic polynomial be
$P(x) =ax^2 +bx +c$
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
$4x^2 + 10x - 6$
$=4x^2 +(-2+12)x -6$
$=4x^2 -2x + 12x -6$
$= 2x(2x-1)+ 6(2x-1)$
$= (2x-1)(2x+6)$
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,
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 $x^{51} +x^2 +1$ Check your Answer