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How to factor polynomials




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 polynomails 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×x) – (7×6)

=7(x-6)
 

ii) 6p-12

=  (6×p) – (6×2)

=6(p-2)

iii) 7a2+14a

=(7×a×a) +  (2×7×a)

=7a(a+2)

iv) -16z+ 20z3

=4z(5z2-4)

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 xy2

=(5×x×x×y)  + (5×3×x×y×y)

=5x(xy-3y2)

 

vii) 10a² -15 b2+20c2

=(2×5×a×a)  - (5×3×b×b) + (5×2×2×c×c)

=5(2a2-3b2+4c2)

 

viii) -4a² +4a b-4ca

=(-4×a×a)  +(4×b×a) - (4×c×a)

=-4a(a-b+c)

 

ix)  x²yz + xy2z  +xyz2

=(x×x×y×z)  +(x×y×y×z) +(x×y×z×z)

=xyz(x+y+z)

x) ax²yz + bxy2z  +cxyz2

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

=xyz(ax+by+cz)

 


factor polynomials by grouping


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) x2 + 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;
x2 + 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,
x2 + 8x - x - 8= x(x+8) -1(x+8)
= (x+8)(x-1)
2) x5 -2x3 -2x2 +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;
x5 -2x3 = x3(x2-2)
What about the last two terms? Observe them. If you take -2 out
-2x2 +4 = -2(x2-2)
Step 3 Putting both part togethes together,
x5 -2x3 -2x2 +4= x3(x2-2) -2(x2-2)
= (x2-2)(x3-2)

factor Quadratic polynomials by splitting the middle term



Let the Quadratic polynomial be
P(x) =ax2 +bx +c
Let its factors be (px + q) and (rx + s). Then
ax2 +bx +c= (px + q) (rx + s) = pr x2 + (ps + qr) x + qs
Comparing the coefficients of x2, 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 ax2 +bx +c, we have to write b as the sum of two numbers whose product is ac
1) 4x2 + 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
4x2 + 10x - 6
=4x2 +(-2+12)x -6
=4x2 -2x + 12x -6
= 2x(2x-1)+ 6(2x-1)
= (2x-1)(2x+6)
2) x5 -2x3 -2x2 +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;
x5 -2x3 = x3(x2-2)
What about the last two terms? Observe them. If you take -2 out
-2x2 +4 = -2(x2-2)
Step 3 Putting both part togethes together,
x5 -2x3 -2x2 +4= x3(x2-2) -2(x2-2)
= (x2-2)(x3-2)

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)

4) Factorising a Polynomial by Factor Theorem



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) 4x2 + 10x - 6
Step 1 Take 4 out of the expression 4( x2 + 5x/2 -3/2)
Step 2 This can be further expressed as
4( x2 + 5x/2 -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
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( x2 + 5x/2 -3/2) = 4(x-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
P(x)= x3 +6x2+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
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
Step 2 x3 +6x2+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
x3 +6x2+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
x3 +6x2+11x+6
= x3 +x2+5x2+5x+6x +6
Splitting cubic polynomial to get (x+1)
=x2(x+1) + 5x(x+1) + 6(x+1)
=(x+1)(x2+5x+6)
Now solving the quadratic polynomial by splitting method
=(x+1)(x2+2x+3x+6)
=(x+1)[x(x+2)+3(x+2)]
=(x+1)(x+2)(X+3)
In General Term,

S(x)=anxn+a(n-1)x(n-1)+a(n-2)x(n-2)+....+ax+a0

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

Factoring polynomial using Algebraic Identities


Identity I : (x + y)2 = x2 + 2xy + y2

Identity II : (x - y)2 = x2 - 2xy + y2

Identity III : x2 - y2 = (x + y) (x - y)

Identity IV : (x + a) (x + b) = x2 + (a + b)x + ab

Identity V :(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

Identity VI : (x + y)3 = x3 + y3 + 3xy (x + y)

Identity VII :(x - y)3 = x3 - y3 - 3xy(x - y)

Identity VIII : x3 + y3 + z3 -3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

Identity IX : x3 + y3 = (x + y )(x2 + y2 - xy )

Identity X : x3 - y3 = (x - y )(x2 + y2 + xy )

1) 27x3 +1
Step 1 Both th terms are exact cubes (3x) and 1
Step 2 We can the algebric identity IX
27x3 +1
=(3x+1)(9x2+1-3x)

2)8x3 + y3 + 27z3 -18xyz
= (2x)3 + y3 + (3z)3 - 3(2x)(y)(3z)
= (2x + y + 3z)[(2x)2 + y2 + (3z)2 - (2x)(y) - (y)(3z) - (2x)(3z)]
=
(2x + y + 3z) (4x2 + y2 + 9z2 - 2xy - 3yz - 6xz)

Solved Examples Polynomials


1) If x + 1/x = 11, evaluate x2 + 1/x2

Solution x + 1/x = 12
(x + 1/x)2 = (x)2 +1/x2 + 2(x)(1/x) = 144
x2 + 1/x2 + 2 = 144
x2+ 1/x2 = 144 - 2 = 142

2) If 9x2+ 25y2 = 181 and xy = -6. Find the value of 3x + 5y

Solution 9x2+ 25y2 = 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 x2 + y2 + x2 -xy -yz -zx is always non-negative for all values of x, y and y.

Solution To prove x2 + y2 + x2 -xy -yz -zx = 0.
We know that square of any number (+ve or -ve) is always +ve.
x2 + y2 + z2 -xy -yz -zx
= (1/2)[2x2 + 2y2 + 2z2 -2xy -2yz -2zx]
=(1/2)[x2 + y2 -2xy + y2 + z2 -2yz + x2 + z2 -2zx]
=(1/2)[(x2 + y2 -2ab) + (y2 + z2 -2yz) + (x2 + z2 -2zx)]
=(1/2)[(x - y)2 + (y - z)2 + (z - x)2 ]
Here, all the terms are always be positive,
x2 + y2 + x2 -xy -yz -zx = 0

Practice Questions


1) If ( x - 4) is a factor of the polynomial 2x2 + Ax + 12 and ( x - 5) is a factor of the polynomial x3 - 7x2+ 11 x + B , then what is the value of ( A - 2B )?
2) if x -1/x =3; then find the value of x3 -1/x3
3) if a+b+c=7 and ab+bc+ca=20 find the value of (a+b+c)2

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