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kiruha [24]
4 years ago
15

4(x-y)^2-12(x-y)(x+y)+9(x+y)^2 ​

Mathematics
2 answers:
tigry1 [53]4 years ago
8 0

Answer:

(x + 5 y)^2

Step-by-step explanation:

Simplify the following:

4 (x - y)^2 - 12 (x - y) (x + y) + 9 (x + y)^2

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

4 x^2 - 2 x y + y^2 - 12 (x - y) (x + y) + 9 (x + y)^2

4 (x^2 - 2 x y + y^2) = 4 x^2 - 8 x y + 4 y^2:

4 x^2 - 8 x y + 4 y^2 - 12 (x - y) (x + y) + 9 (x + y)^2

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

4 x^2 - 8 x y + 4 y^2 - 12 x^2 - y^2 + 9 (x + y)^2

-12 (x^2 - y^2) = 12 y^2 - 12 x^2:

4 x^2 - 8 x y + 4 y^2 + 12 y^2 - 12 x^2 + 9 (x + y)^2

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

4 x^2 - 8 x y + 4 y^2 - 12 x^2 + 12 y^2 + 9 x^2 + 2 x y + y^2

Grouping like terms, 4 x^2 - 8 x y + 4 y^2 - 12 x^2 + 12 y^2 + 9 (x^2 + 2 x y + y^2) = 9 (x^2 + 2 x y + y^2) + (4 y^2 + 12 y^2) - 8 x y + (4 x^2 - 12 x^2):

9 (x^2 + 2 x y + y^2) + (4 y^2 + 12 y^2) - 8 x y + (4 x^2 - 12 x^2)

4 y^2 + 12 y^2 = 16 y^2:

9 (x^2 + 2 x y + y^2) + 16 y^2 - 8 x y + (4 x^2 - 12 x^2)

4 x^2 - 12 x^2 = -8 x^2:

9 (x^2 + 2 x y + y^2) + 16 y^2 - 8 x y + -8 x^2

9 (x^2 + 2 x y + y^2) = 9 x^2 + 18 x y + 9 y^2:

9 x^2 + 18 x y + 9 y^2 + 16 y^2 - 8 x y - 8 x^2

Grouping like terms, 9 x^2 + 18 x y + 9 y^2 + 16 y^2 - 8 x y - 8 x^2 = (9 y^2 + 16 y^2) + (18 x y - 8 x y) + (9 x^2 - 8 x^2):

(9 y^2 + 16 y^2) + (18 x y - 8 x y) + (9 x^2 - 8 x^2)

9 y^2 + 16 y^2 = 25 y^2:

25 y^2 + (18 x y - 8 x y) + (9 x^2 - 8 x^2)

x y 18 + x y (-8) = 10 x y:

25 y^2 + 10 x y + (9 x^2 - 8 x^2)

9 x^2 - 8 x^2 = x^2:

25 y^2 + 10 x y + x^2

The factors of 25 that sum to 10 are 5 and 5. So, 25 y^2 + 10 x y + x^2 = (x + 5 y) (x + 5 y):

(x + 5 y) (x + 5 y)

(x + 5 y) (x + 5 y) = (x + 5 y)^2:

Answer:  (x + 5 y)^2

marysya [2.9K]4 years ago
4 0

4(x-y)^2-12(x-y)(x+y)+9(x+y)^2 =\\4(x^2-2xy+y^2)-12(x^2-y^2)+9(x^2+2xy+y^2)=\\4x^2-8xy+4y^2-12x^2+12y^2+9x^2+18xy+9y^2=\\x^2+10xy+25y^2

which can factorised into (x+5y)^2

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Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

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We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

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3 years ago
Complete the square to rewrite y-x^2-6x+14 in vertex form. then state whether the vertex is a maximum or minimum and give its co
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Answer:

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And solving we have:

y= x^2 -6x +9 + 14 -9

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And we can write the expression like this:

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The vertex for this case would be:

V= (3,5)

And the minimum for the function would be 3 and there is no maximum value for the function

Step-by-step explanation:

For this case we have the following equation given:

y= x^2 -6x +14

We can complete the square like this:

y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2

And solving we have:

y= x^2 -6x +9 + 14 -9

y= (x-3)^2 +5

And we can write the expression like this:

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The vertex for this case would be:

V= (3,5)

And the minimum for the function would be 3 and there is no maximum value for the function

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