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Try this hard Math Problem if you dare!!

Try this hard Math Problem if you dare!!-example-1
User Max Meijer
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1 Answer

4 votes

Answer:

a.
(x - 3)^2 + 16

b.
8(x -7)^2

c.
(a^2 - 1)(7x - 6) or
(a+1)(a-1)(7x-6)

d.
(x^2-4)(x^2+3) or
(x-2)(x+2)(x^2+3)

e.
(a^n+b^n)(a^n-b^n)(a^(2n) +b^(2n))

Explanation:


a.\ (x + 1)^2 - 8(x - 1) + 16

Expand


(x + 1)(x + 1) - 8(x - 1) + 16

Open brackets


x^2 + x + x + 1 - 8x + 8 + 16


x^2 + 2x + 1 - 8x + 24

Collect Like Terms


x^2 + 2x - 8x+ 1 + 24


x^2 - 6x+ 25

Express 25 as 9 + 16


x^2 - 6x+ 9 + 16

Factorize:


x^2 - 3x - 3x + 9 + 16


x(x -3)-3(x - 3) + 16


(x - 3)(x - 3) + 16


(x - 3)^2 + 16


b.\ 8(x - 3)^2 - 64(x-3) + 128

Expand


8(x - 3)(x - 3) - 64(x-3) + 128


8(x^2 - 6x+ 9) - 64(x-3) + 128

Open Brackets


8x^2 - 48x+ 72 - 64x+192 + 128

Collect Like Terms


8x^2 - 48x - 64x+192 + 128+ 72


8x^2 -112x+392

Factorize


8(x^2 -14x+49)

Expand the expression in bracket


8(x^2 -7x-7x+49)

Factorize:


8(x(x -7)-7(x-7))


8((x -7)(x-7))


8(x -7)^2


c.\ 7a^2x - 6a^2 - 7x + 6

Factorize


a^2(7x - 6) -1( 7x - 6)


(a^2 - 1)(7x - 6)

The answer can be in this form of further expanded as follows:


(a^2 - 1^2)(7x - 6)

Apply difference of two squares


(a+1)(a-1)(7x-6)


d.\ x^4 - x^2 - 12

Express
x^4 as
x^2


(x^2)^2 - x^2 - 12

Expand


(x^2)^2 +3x^2- 4x^2 - 12


x^2(x^2+3) -4(x^2+3)


(x^2-4)(x^2+3)

The answer can be in this form of further expanded as follows:


(x^2-2^2)(x^2+3)

Apply difference of two squares


(x-2)(x+2)(x^2+3)


e.\ a^(4n) -b^(4n)

Represent as squares


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

Apply difference of two squares


(a^(2n) -b^(2n))(a^(2n) +b^(2n))

Represent as squares


((a^(n))^2 -(b^(n))^2)(a^(2n) +b^(2n))

Apply difference of two squares


(a^n+b^n)(a^n-b^n)(a^(2n) +b^(2n))

User Unify
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