- 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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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 polynomails are

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×x) – (7×6)

=7(x-6)

ii) 6p-12

= (6×p) – (6×2)

=6(p-2)

iii) 7a^{2}+14a

=(7×a×a) + (2×7×a)

=7a(a+2)

iv) -16z+ 20z^{3}

=4z(5z^{2}-4)

v) 20 l² m + 30 a l m

=(2×2×5×l×m×l) + (2×3×5×a×l×m)

=2lm(10l+15a)

vi) 5 x²y -15 xy^{2}

=(5×x×x×y) + (5×3×x×y×y)

=5x(xy-3y^{2})

vii) 10a² -15 b^{2}+20c^{2}

=(2×5×a×a) - (5×3×b×b) + (5×2×2×c×c)

=5(2a^{2}-3b^{2}+4c^{2})

viii) -4a² +4a b-4ca

=(-4×a×a) +(4×b×a) - (4×c×a)

=-4a(a-b+c)

ix) x²yz + xy^{2}z +xyz^{2}

=(x×x×y×z) +(x×y×y×z) +(x×y×z×z)

=xyz(x+y+z)

x) ax²yz + bxy^{2}z +cxyz^{2}

=(x×x×y×z×a) +(x×y×y×z×b) +(x×y×z×z×c)

=xyz(ax+by+cz)

When we dont 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^{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 togethes 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 togethes together,

x^{5} -2x^{3} -2x^{2} +4= x^{3}(x^{2}-2) -2(x^{2}-2)

= (x^{2}-2)(x^{3}-2)

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

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

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

Step 3 Putting both part togethes together,

x

= (x+8)(x-1)

2) x

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

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

-2x

Step 3 Putting both part togethes together,

x

= (x

Let the Quadratic polynomial be

P(x) =ax

Let its factors be (px + q) and (rx + s). Then

ax

Comparing the coefficients of x

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

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)

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 togethes together,

x^{5} -2x^{3} -2x^{2} +4= x^{3}(x^{2}-2) -2(x^{2}-2)

= (x^{2}-2)(x^{3}-2)

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

=4x

=4x

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

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

2) x

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

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

-2x

Step 3 Putting both part togethes together,

x

= (x

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 know by factor theorem if (x-a) is the factor of the polynomial ,then P(a)=0. We can use this theorem to find factors of the polynomials

Lets first take a look at quadratic polynomial

1) 4x

Step 1 Take 4 out of the expression 4( x

Step 2 This can be further expressed as

4( x

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

Whereever 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

= (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

P(x)= x

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

Whereever it satisfies the factor theorem, we are good

In this particular case

P(-1)=P(-2)=P(-3)=0, we can write like this

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

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

= x

Splitting cubic polynomial to get (x+1)

=x

=(x+1)(x

Now solving the quadratic polynomial by splitting method

=(x+1)(x

=(x+1)[x(x+2)+3(x+2)]

=(x+1)(x+2)(X+3)

In General Term,

S(x)=a

Look for the factors in a

**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 algebric 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)

Step 1 Both th terms are exact cubes (3x) and 1

Step 2 We can the algebric identity IX

27x

=(3x+1)(9x

2)8x

= (2x)

= (2x + y + 3z)[(2x)

=

(2x + y + 3z) (4x

1) If x + 1/x = 11, evaluate x^{2} + 1/x^{2}

**Solution **
x + 1/x = 12

(x + 1/x)^{2} = (x)^{2} +1/x^{2} + 2(x)(1/x) = 144

x^{2} + 1/x^{2} + 2 = 144

x^{2}+ 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 y.

**Solution **
To prove x^{2} + y^{2} + x^{2} -xy -yz -zx = 0.

We know that square of any number (+ve or -ve) is always +ve.

x^{2} + y^{2} + z^{2} -xy -yz -zx

= (1/2)[2x^{2} + 2y^{2} + 2z^{2} -2xy -2yz -2zx]

=(1/2)[x^{2} + y^{2} -2xy + y^{2} + z^{2} -2yz + x^{2} + z^{2} -2zx]

=(1/2)[(x^{2} + y^{2} -2ab) + (y^{2} + z^{2} -2yz) + (x^{2} + z^{2} -2zx)]

=(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 = 0

(x + 1/x)

x

x

2) If 9x

(3x)

(3x)

(3x)

(3x + y)

(3x + y)

3x + y = +1 or -1

3) Prove that x

We know that square of any number (+ve or -ve) is always +ve.

x

= (1/2)[2x

=(1/2)[x

=(1/2)[(x

=(1/2)[(x - y)

Here, all the terms are always be positive,

x

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 -1/x =3; then find the value of x^{3} -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 -1/x =3; then find the value of x

3) if a+b+c=7 and ab+bc+ca=20 find the value of (a+b+c)

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