Pls help me with this equation, we must use grouping to get the answer: m^2 - 6mn + 9n^2 - 4a^2 + 4ab - b^2    (the sign "^" refers to the exponent)heres an example:  x^4 -(4x-5)^2 the answer is...

Pls help me with this equation, we must use grouping to get the answer:

m^2 - 6mn + 9n^2 - 4a^2 + 4ab - b^2    (the sign "^" refers to the exponent)

heres an example:  x^4 -(4x-5)^2 the answer is [x^2 + (4x-5)] [x^2 -(4x-5)]  and so on...

Asked on by red311

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samhouston's profile pic

samhouston | Middle School Teacher | (Level 1) Associate Educator

Posted on

m^2 - 6mn + 9n^2 - 4a^2 - 4ab + b^2

Separate this equation into 2 polynomials using grouping symbols.

(m^2 - 6mn + 9n^2) - (4a^2 - 4ab + b^2)

Factor each polynomial.  Both polynomials are Perfect Squares.

(m - 3n)^2 - (2a - b)^2

If you consider (m - 3n) as x and (2a - b) as y, then you get a simpler equation:

x^2 - y^2

This is called the Difference of Squares.  According to the difference of squares...

x^2 - y^2 = (x + y)(x - y)

Now replace (m - 3n) for x and (2a - b) for y.  Remember the - is distributed in the second set of (2a - b).

Answer:  (m - 3n + 2a - b)(m - 3n - 2a + b)

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll create the following groups:

(m^2 - 6mn + 9n^2) - (4a^2 - 4ab + b^2)

We notice that within brackets, we'll have two pwerfect squares that return the formula:

x^2 - 2xy + y^2 = (x - y)^2

We'll manage the 1st pair of brackets:

Let x = m and y = 3n => m^2 - 6mn + 9n^2 = (m - 3n)^2

We'll manage the 2nd pair of brackets:

Let x = 2a and y = b => 4a^2 - 4ab + b^2 = (2a - b)^2

The difference between squares will be:

(m - 3n)^2 - (2a - b)^2

But we know that the difference between two squares returns the product:

x^2 - y^2 = (x-y)(x+y)

(m - 3n)^2 - (2a - b)^2 = (m - 3n - 2a + b)(m - 3n + 2a - b)

Therefore, the equivalent product of the given expression is: (m - 3n)^2 - (2a - b)^2 = (m - 3n - 2a + b)(m - 3n + 2a - b).

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