Factorization Of Algebraic Expressions

Given,

Given,

use the identity:

(a + b)² = a² + b² + 2ab

Given,

As (x³ - y³) = (x – y)(x² + xy + y²)

x(x³–y³)+3xy-(x–y) = x [(x – y)(x² + xy + y²)] + 3xy-(x–y)

Take x(x-y) common to get,

x(x-y)[ (x² + xy + y²) + 3y]

Given,

Given,

Given,

Given,

6ab-b²+12ac-2bc = b(6a-b)+2c(6a-b)

= (b+2c)(6a-b)

Given,

[ BY applying (a² - 2ab + b² ) = (a-b)² ]

Given,

X(x-2) (x-4)+4x-8 = x(x-2) (x-4)+4(x-2)

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

= (x-2) (x² - 4x + 4)

= (x-2) (x-2)²

= (x-2)³

Given,

(x+2) (x²+25) - 10x² - 20x = (x+2) (x²+25)-10x (x+2)

= (x+2) (x²+25-10x)

= (x+2) (x-5)²

Given,

2a²√6ab+3b² = (√2a)² +2(√3×√2)ab + (√3b)²

= (√2a+√3b)²

Given,

(a-b+c)²+(b-c+a)²+2(a-b+c)(b-c+a)

= (a-(b-c))² + (a+(b-c))² + 2(a-(b-c)) (a+(b-c))

[Applying identity: x² + y² +2xy = (x + y)² , where x= a-(b-c), y= a+(b-c)]

Given,

a²+b²+2(ab+bc+ca)

= a²+b²+2ab+2bc+2ca

= (a+b)²+2c(a+b)

= (a+b)(a+b+2c)

Given,

4(x-y)² -12(x-y)(x+y)+9(x+y)² = 4(x²-2xy+y²)-12(x² – y²)+9(x²+y²+2xy)

= 4x²-8xy+4y²-12x²+12y²+9x²+9y+18xy

= x²+25y²+10xy

= (x)²+(5y)²+2×x×5y

= (x+5y)²

Given,

a²-b²+2ab-c² = a²-(b²-2bc+c²)

= a – (b-c)²

= (a + (b-c)) (a-(b-c))

= (a+b-c) (a-b+c)

Given,

a²+2ab+b²-c² = (a+b)² – c²

= (a+b-c) (a+b+c)

Given,

a² +4b²-4ab-4c² = (a)²+ (2b)² – 2×a×2b-4c²

= (a-2b)² – (2c)²

= (a-2b-2c)(a-2b+2c)

Given,

xy⁹ - yx⁹ = xy (y⁸-x⁸)

= xy ((y⁴)² – (x⁴)²)

= xy (y⁴+x⁴) (y⁴ - x⁴)

= xy (y⁴+x⁴) (y²-x²) (y²+x²)

= xy (y⁴+x⁴) (y²+x²) (y – x) (y+x)

Given,

x⁴+x²y²+y⁴ = x⁴+2x²y²+y⁴ – x²y²

= (x²y²)² – (xy)²

= (x²+ y² – xy) (x²+ y² + xy)

Given,

x² – y² - 4xz + 4z² = x² – 4xz +4z² – y²

= (x)² - 2×x×2z+(2z)² – y²

= (x-2z)² – y²

= (x - 2z - y) (x – 2z + y)

= (x – y – 2z) (x + y – 2z)

Given,

x²+6√2x+10 = x²+√2x + 10

= x(x+√2) + 5√2 (x +√2)

= (x+√2) (x+5√2)

Given,

x² - 2√2x – 30 = x² - 5√2x+3√2x-30

= x(x-5√2) +3√2 (x - 5√2)

= (x-5√2) (x+3√2)

Given,

X² - √3x – 6 = x² - 2√3x + √3x – 6

= x (x - 2√3) + √3 (x-2√3)

= (x + √3) (x - 2√3)

Given,

x² + 5√5x + 30 = x² + 3√5x + 2√5x + 30

= x (x + 3√5) + 2√5 (x + 3√5)

= (x + 3√5) (x + 2√5)

Given,

x² + 2√3x – 24 = x² + 4√3x - 2√3x – 24

= x(x + 4√3) - 2√3 (x + 4√3)

= (x + 4√3) (x - 2√3)

Given,

Given,

Given,

Given,

5√5x²+20x+3√5 = 5√5x² + 15x + 5x + 3√5

= 5x(√5x+3) + √5 (√5x+3)

= (5x +√5) (√5x + 3)

Given,

2x²+3√5x+5 = 2x² + 2√5x+ √5x+5

= 2x (x + √5) +√5 (x+√5)

= (2x+√5) (x+√5)

Given,

9(2a – b)² –4 (2a – b) – 13

Let us assume (2a – b) = x

9x² – 4x – 13

9x² – 13x + 9x – 13

x(9x – 13)+1 (9x + 3)

(9x – 13) (x + 1)

[9(2a – b) – 13] [2a – b + 1]

(18a – 9b – 13) (2a – b + 1)

Given,

7(x – 2y)² – 25 (x – 2y) + 12

Let a = (x – 2y),

So we have,

= 7a² – 25a + 12

= 7a² – 21a – 4a + 12

= 7a(a - 3) -4 (a - 3)

= (7a – 4) (a - 3)

Put a = (x – 2y)

= {7(x – 2y) – 4} (x – 2y - 3)

= (7x – 14y -4) (x – 2y - 3)

Given,

2(x+y)² – 9(x+y) – 5,

= 2a² -9a – 5

Let (x+y) = a

= 2a² – 10a+a – 5

= 2a (a-5) +1(a-5)

= (2a+1) (a-5)

= {2(x+y)+1}(x+y-5)

(2x+2y+1) (x+y-5)

We know that,

Area of rectangle = length × breadth

Given,

35y² + 13y – 12 = 35y² + 28y – 15y -12

= 7y(5y+4) -3 (5y + 4)

= (7y – 3) (5y + 4)

Thus,

Length = (7y – 3), then breadth = (5y + 4)

Length = (5y + 4), then breadth = (7y – 3)

Given,

We know that,

Volume of cuboids = length × breadth × height

Given,

3x² – 12x = 3x (x-4)

Thus,

Dimensions of cuboids are –

Given,

P³+27,

= p³ + (3)³ [∵ a³+b³ = (a+b)(a² – 2ab +b²)

= (p + 3) (p² + 9 – 3p)

Given,

y³+125,

= y³ + (5)³ [∵ a³+b³ = (a+b)(a² – 2ab +b²)

= (y + 5) (y² – 5y + 25)

Given,

1 – 27a³,

= 1 – (3a)³

= (1 – 3a) (1+9a²+3a)

Given,

8x³ y³ +27a³,

= (2xy)³ + (3a)³

= (2xy + 3a) (4x² y² +9a² – 6axy)

Given,

64a³ - b³,

= (4a)³ - (b)³

= (4a - b) (16a² + b² + 4ab)

Given,

Given,

10x⁴ y – 10xy⁴,

= 10xy (x³ - y³)

= 10xy (x - y) (x² + xy – y²)

Given,

54x⁶ y + 2x³ y⁴,

= 2x³ y (27x³ + y³)

= 2x³ y {(3x)³ + (y)³}

= 2x³ y (3x + y) (9x² + y² – 3xy)

Given,

32a³+108b³,

= 4 (8a³+27b³)

= 4 { (2a)³ + (3b)³ }

= 4 (2a + 3b) (4a² +9b² – 6ab)

Given,

(a – 2b)³ – 512b³

= (a – 2b)³ – (8b)³

= (a – 2b – 8b) {(a – 2b)² +(8b)² +(a – 2b) 8b}

= (a – 10b) (a² +4b² – 4ab +64b² +8ab – 16b²)

= (a – 10b) (a² +52b² +4ab)

Given,

(a + b)³ – {2(a – b)}³

= {(a + b) –2(a - b)} { (a+b)² +4(a-b)² +2(a+b)(a - b)} [ By using: x³ - y³ = (x - y)(x² + y² +xy)

= (a + b – 2a + 2b)(a² + b² + 2ab + 4a² +4b² – 8ab + 2a² – 2b²)

= (3b – a) (7a² +3b² - 6ab)

Given,

(x+2)³ + (x - 2)³ = (x + 2 + x - 2) { (x + 2)² + (x - 2)² – (x + 2)(x - 2) }

= 2x (x² + 4 + 4x + x² + 4 – 4x – x² + 4)

= 2x (x² +12)

Given,

8x²y³ – x⁵

= x² (8y³ – x³)

= x² { (2y)² – (x)³}

= x² (2y - x) (4y² + x² + 2xy)

Given,

1029 – 3x³

= 3 (343 – x³)

= 3 { (7)³ – (x)³}

= 3 (7 - x) (49 +x² +7x)

Given,

X⁶ + y⁶ = (x²)³ + (y²)³

= (x² + y²) (x⁴ +y⁴ – x²y²)

Given,

X³y³ + 1 = (xy)³ +(1)³

= (xy + 1) (x² y² +1 - xy)

Given,

X⁴y⁴ – xy = xy (x³y³ -1)

= xy { (xy)³ – (1)³}

= xy (xy -1) (x²y² +1+xy)

Given,

a¹² +b¹² = (a⁴)³ + (b⁴)³

= (a⁴ + b⁴)(a³ +b³ – a⁴b⁴)

Given,

X³ +6x²12x+16 = (x³ + 6x²+12x+8)+8

= (x+2)³ + 8 [∵ (a+b)³ = a³ + b³ +3ab(a+b)]

= (x+2)³ + (2)³

= (x+2+2) {(x+2)² +4 – 2 (x+2)}

= (x+4)(x²+4+4x+4 - 2x - 4)

= (x+4)(x²+2x+4)

Given,

a³+b³+a+b = (a³ + b³) + (a+b)

= (a+b) (a² – ab +b²) +a+b

= (a+b)(a² – ab +b²+1)

Given,

Given,

A³+3a² b+3ab²+b³-8 = (a³ +3a²b +3ab²+b³) – 8

= (a+b)³ -8

= (a+b)³ – (2)³

= (a+b -2) { (a+b)² + (2)² + 2(a+b) }

= (a+b - 2) (a²+b² + 2ab+4+2a+2b)

Given,

8a³ – b³ – 4 ax + 2bx = (2a)³ – (b)³ – 2x(2a - b)

= (2a - b) (4a² + b² +2ab)-2x(2a - b)

= (2a - b) (4a² + b² +2ab – 2x)

(i) Given,

(ii) Given,

(ii) Given,

=

Given,

64a³ + 125b³ + 240a²b + 300ab²,

= (4a)³ + (5b)³ + 60ab (4a + 5b)

= (4a)³ + (5b)³ + 3×4a×5b (4a + 5b)

= (4a+5b)³

= (4a+5b)(4a+5b)(4a+5b)

Given,

125x³-27y³-225x²y+125xy²,

= (5x)³ – (3y)³ – 45xy (5x – 3y)

= (5x)³ – (3y)³ – 3×5x×3y (5x – 3y)

= (5x – 3y)³

= (5x – 3y) (5x – 3y) (5x – 3y)

Given,

Given,

8x³+27y³+36x²y+54xy²,

= (2x)³+(3y)³ +18xy (2x +3y)

= (2x)³+(3y)³ +3×2x×3y (2x +3y)

= (2x+3y)³

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

Given,

a³ - 3a²b + 3ab² - b³ + 8,

= {(a)³ – (b)³ -3ab (a-b)} +8

= (a-b)³ + (2)³

= (a – b +2) {(a - b)² +(2)² +2(a - b)}

= (a - b +2)(a² -2ab +b² +4+2a – 2b)

Given,

x³+8y³+6x²y+12xy²,

= (x)³+(2y)³ +6xy (x +2y)

= (x)³+(2y)³ +3×x×2y (x +2y)

= (x+2y)³

= (x +2y) (x +2y) (x +2y)

Given,

8x³+y³+12x²y+6xy²,

= (2x)³+(y)³ +6xy (2x +y)

= (2x)³+(y)³ +3×2x×y (2x +y)

= (2x+y)³

= (2x +y) (2x +y) (2x +y)

Given,

8a³+27b³+36a²b+54ab²,

= (2a)³+(3b)³ +18ab (2a + 3b)

= (2a)³ + (3b)³ +3×2a×3b (2a +3b)

= (2a +3b)³

= (2a +3b) (2a + 3b) (2a +3b)

Given,

8a³ - 27b³ - 36a²b+54ab²,

= (2a)³ - (3b)³ - 18ab (2a - 3b)

= (2a)³ - (3b)³ - 3×2a×3b (2a – 3b)

= (2a - 3b)³

= (2a - 3b) (2a - 3b) (2a - 3b)

Given,

x³-12x(x-4)-64,

= (x)³ -12x (x - 4) – (4)³

= (x)³ –(4)³ - 3×x×4 (x - 4)

= (x - 4)³

= (x - 4) (x - 4) (x - 4)

Given,

a³x³-3a²bx²+3ab²x-b³,

= (ax)³ – 3abx(ax-b) – (b)³

= (ax)³ – (b)³ – 3abx (ax-b)

= (ax-b)³

= (ax-b) (ax-b) (ax-b)

Given,

=

This can be written in form

=

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Hence,

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Thus the required factors of

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Thus the required factors of

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Thus the factors of

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Thus the factors of

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Thus the factors of

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Thus the factors of

Given,

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Let

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Here ,

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Hence ,

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= = 3

Given,

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Let

Then ,

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Here ,

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Hence ,

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= .

Given,

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Let

Then ,

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Here ,

= = 0

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Hence ,

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Let

Then,

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Here ,

= = 0

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Hence,

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This can be written in form ,

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Hence,

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Hence ,

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Thus the factors of

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Hence ,

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Thus the factors of

Given,

=

= x + y = -4 Given

= x+y+4 = 0

This can be written in form ,

=

And ,

=

= 0 ×

= 0

(i) Given,

=

Multiply the above expression by ( x+ y – z)

=

=

=

=

(ii) Given,

=

Multiply above expression by (x-2y-z)

Then ,

=

=

By formula…

=

iii) we have

=

(iii) Given,

=

Multiply above equation by ( x- 2y +3 )

=

=

=

=

(iv) Given,

=

Multiply above equation by (3x – 5y +4)

We got,

=

=

=

=

We have,

=

=

=

Thus, the factors of

We have,

First we rewrite the question,

x⁴ + x² + 25 = (x²)² + 2.x².5 + 52 - 9x²

= {(x²)² + 2.x².5 + 52} – (3x)² [By using a² + 2ab + b² = (a + b)²]

= {x² + 5}² – (3x)² [ By using a² – b² = (a + b) (a - b)

= (x² + 5 +3x) ( x² + 5 -3x)

Thus , the factors of .

We have,

=

Adding and subtracting 9x² in the equation

=

=

=

Thus the factors of (x⁴+x²+25) are

We have ,

=

Taking -1 as common from last three terms

=

=

=

=

Thus the factors of

We have,

=

When ( a + b + c) = 0 Given

=

=

=

We have ,

=

=

Let a = ( x-2y) , then the expression becomes ,

=

=

= a(a-4) + 2( a -4)

=

Put a = ( x -2y)

=

Thus the factors of

We have,

=

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As

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We have ,

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Then,

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We have ,

=

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Thus the factors of

We have ,

=

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Then,

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We have,

=

=

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Then,

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We have ,

=

=

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Thus the factors of are .

We have ,

=

Applying formulas,

=

=

=

=

Thus the factors of are

We have,

=

Let a = 25 , b = -75 , c = 50 ,

Then the expression becomes as ,

=

=

Here ,

Hence,

=

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=

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=

We have,

=

Let a = 48 , b = -30 , c = -18

Then the expression becomes ,

=

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Here,

=

Hence,

=

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We have,

=

Adding and subtracting 1 in the equation

=

=

=

=

Thus the factors of

We have ,

=

Let

=

Here,

=

Hence ,

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We have,

Let

So,

If a+b+c = 0 , then,

=

=

We have ,

=

Let

=

Here ,

=

Hence,

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=

We have ,

=

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=

We have,

= 3x = a+b+c

Let

So,

= [ a +b + c = 3x ] given

=

Now,

We have,

=

=

=

=

= k = 8 .

We have,

=

=

=

By compairing both sides ,

= a = 1

= a + b = -3

= b + c = 3

= c = -7

Thus , a + (b + c) = 1+3 = 4.

We have,

=

=[

Hence,

=

= ( 2.3 – 0.3 ) = 2 .

We have,

= [

Hence,

=

= 0.013 + 0.007 = 0.020