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College Algebra

by Rey
(Deliman Quezon City, Phillipines)











































(2g+h)^7
(6x-7y)^2
(m/3+m/9)^2
(4x-5y+z)^2
(2r-4b)^3

Comments for College Algebra

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Jul 15, 2011
Binomial & Trinomial Expansion
by: Staff


Part II

------------------------------------------------------------

Before going on, notice the pattern that has developed:


Multiply the first two factors: (2g+h)^2

= (2g)² + 2*(2g)¹*h + h²)

The 3 coefficients 1, 2, 1 are the third row of Pascal’s Triangle



Multiply the first three factors: (2g+h)^3

= (2g)³ + 3*(2g)²*h+ 3*2g*h² + h³

The 4 coefficients 1, 3, 3, 1 are the fourth row of Pascal’s Triangle




Multiply the first four factors: (2g+h)^4

= (2g)⁴ + 4*(2g)³*h + 6*(2g)²*h² + 4*(2g)*h³ + h⁴

The 5 coefficients 1, 4, 6, 4, 1 are the fifth row of Pascal’s Triangle



Multiply the first five factors: (2g+h)^5

= (2g)⁵ + 5*(2g)⁴*h + 10*(2g)³*h² + 10*(2g)²*h³ + 5*(2g)*h⁴ + h⁵

The 6 coefficients 1, 5, 10, 10, 5, 1 are the sixth row of Pascal’s Triangle



Multiply the first six factors: (2g+h)^6

= (2g)⁶ + 6*(2g)⁵*h + 15*(2g)⁴*h² + 20*(2g)³*h³ + 15*(2g)²*h⁴ + 6*(2g)*h⁵ + h⁶

The 7 coefficients 1, 6, 15, 20, 15, 6, 1 are the seventh row of Pascal’s Triangle



Multiply all seven factors: (2g+h)^7

= (2g)⁷ + 7*(2g)⁶* h + 21*(2g)⁵*h² + 35*(2g)⁴*h³ + 35*(2g)³*h⁴ + 21*(2g)²*h⁵ + 7*(2g)*h⁶ + h⁷

The 8 coefficients 1, 7, 21, 35, 35, 21, 7, 1 are the eighth row of Pascal’s Triangle


We can use this pattern to solve the problems part B. and E.

------------------------------------------------------------


B. (6x-7y)^2

The expression (6x-7y) is multiplied by itself two times (because the exponent is ^2)

= (6x-7y)*(6x-7y)

= (6x)² + 2*(6x)¹*(-7y) + (-7y)²

= 36x² - 84xy + 49y²

The final answer to part B: (6x-7y)^2

= 36x² - 84xy + 49y²



C. (m/3+m/9)^2

To simplify this problem, combine the two fractions:

(m/3+m/9)^2

= (3m/9 + m/9)^2

= [(3m + m)/9]^2

= (4m/9)^2





The expression (4m/9) is multiplied by itself two times (because the exponent is ^2)

= (4m/9)*(4m/9)

= (16m²)/81

The final answer to part C: (m/3+m/9)^2

= (16m²)/81



D. (4x-5y+z)^2

The expression (4x-5y+z) is multiplied by itself two times (because the exponent is ^2)

= (4x-5y+z)*(4x-5y+z)

= 4x*4x + 4x*(-5y) + 4x*z + (-5y)*4x + (-5y)*(-5y) + (-5y)*z + z*4x + z*(-5y) + z*z

= 16x² - 20xy + 4xz – 20xy + 25y – 5yz + 4xz – 5yz + z²

= 16x² - 40xy + 8xz + 25y² - 10yz + z²

The final answer to part D: (4x-5y+z)^2

= 16x² - 40xy + 8xz + 25y² - 10yz + z²




E. (2r-4b)^3


The expression (2r-4b) is multiplied by itself three times (because the exponent is ^3)

= (2r-4b)*(2r-4b)*(2r-4b)

= (2r)³ + 3*(2r)²*(-4b) + 3*2r *(-4b)² + (-4b)³

= 8r³ - 48br² + 96b²r - 64b³





The final answer to part E: (2r-4b)^3

= 8r³ - 48br² + 96b²r - 64b³





Thanks for writing.

Staff
www.solving-math-problems.com




Jul 15, 2011
Binomial & Trinomial Expansion
by: Staff


Part I

The question:

by Rey
(Deliman Quezon City, Phillipines)

(2g+h)^7
(6x-7y)^2
(m/3+m/9)^2
(4x-5y+z)^2
(2r-4b)^3

The answer:

A. (2g+h)^7
B. (6x-7y)^2
C. (m/3+m/9)^2
D. (4x-5y+z)^2
E. (2r-4b)^3



The fastest way to expand binomial expressions is to use Pascal’s triangle to calculate the coefficients.


But first, we’ll complete the expansion by hand for part A.


(2g+h)^7

The expression (2g+h) is multiplied by itself seven times (because the exponent is ^7)

= (2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)


Multiply 1st two factors

= [(2g+h)*(2g+h)]*(2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)

= [(2g)² + 2*(2g)¹*h + h²]*(2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)

= (4g² + 4gh + h²)*(2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)


Multiply the result by the 3rd factor

= (2²g² + 4gh + h²)*(2g+h)*(2g+h)*(2g+h)*(2g+h)*(2g+h)

= [(2²g² + 4gh + h²)*(2g+h)]*(2g+h)*(2g+h)*(2g+h)*(2g+h)

= [(2g)³ + 3*(2g)²*h+ 3*2g*h² + h³]*(2g+h)*(2g+h)*(2g+h)*(2g+h)


= (8g³ + 12g²h + 6gh² + h³)*(2g+h)*(2g+h)*(2g+h)*(2g+h)


Multiply the result by the 4th factor

= (2³g³ + 12g²h + 6gh² + h³)*(2g+h)*(2g+h)*(2g+h)*(2g+h)

= [(8g³ + 12g²h + 6gh² + h³)*(2g+h)]*(2g+h)*(2g+h)*(2g+h)

= [(2g)⁴ + 4*(2g)³*h + 6*(2g)²*h² + 4*(2g)*h³ + h⁴]*(2g+h)*(2g+h)*(2g+h)

= (16g⁴ + 32g³h + 24g²h² + 8gh³ + h⁴)*(2g+h)*(2g+h)*(2g+h)



Multiply the result by the 5th factor

= (16g⁴ + 32g³h + 24g²h² + 8gh³ + h⁴)*(2g+h)*(2g+h)*(2g+h)

= [(16g⁴ + 32g³h + 24g²h² + 8gh³ + h⁴)*(2g+h)]*(2g+h)*(2g+h)

= [(2g)⁵ + 5*(2g)⁴*h + 10*(2g)³*h² + 10*(2g)²*h³ + 5*(2g)*h⁴ + h⁵]*(2g+h)*(2g+h)

= (32g⁵ + 80g⁴h + 80g³h² + 40g²h³ + 10gh⁴ + h⁵)*(2g+h)*(2g+h)


Multiply the result by the 6th factor

= (32g⁵ + 80g⁴h + 80g³h² + 40g²h³ + 10gh⁴ + h⁵)*(2g+h)*(2g+h)

= [(32g⁵ + 80g⁴h + 80g³h² + 40g²h³ + 10gh⁴ + h⁵)*(2g+h)]*(2g+h)

= [(2g)⁶ + 6*(2g)⁵*h + 15*(2g)⁴*h² + 20*(2g)³*h³ + 15*(2g)²*h⁴ + 6*(2g)*h⁵ + h⁶]*(2g+h)

= (64g⁶ + 192g⁵*h + 240g⁴h² + 160g³h³ + 60g²h⁴ + 12gh⁵ + h⁶)*(2g+h)


Multiply the result by the 7th factor

= (64g⁶ + 192g⁵*h + 240g⁴h² + 160g³h³ + 60g²h⁴ + 12gh⁵ + h⁶)*(2g+h)

= (2g)⁷ + 7*(2g)⁶* h + 21*(2g)⁵*h² + 35*(2g)⁴*h³ + 35*(2g)³*h⁴ + 21*(2g)²*h⁵ + 7*(2g)*h⁶ + h⁷

= 128g⁷ + 448g⁶h + 672g⁵h² + 560g⁴h³ + 280g³h⁴ + 84g²h⁵ + 14gh⁶ + h⁷

The final answer to part A: (2g+h)^7

= 128g⁷ + 448g⁶h + 672g⁵h² + 560g⁴h³ + 280g³h⁴ + 84g²h⁵ + 14gh⁶ + h⁷


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