Question

factorize
9x4 – 34x2y2 +25y4
the 4 and 2 ae squares

Answer 1

Answer:

(x - y) (x + y) (3 x - 5 y) (3 x + 5 y)

Step-by-step explanation:

Simplify the following:

9 x^4 - 34 x^2 y^2 + 25 y^4

9 x^4 - 34 x^2 y^2 + 25 y^4 = 9 (x^2)^2 - 34 x^2 y^2 + 25 (y^2)^2:

9 (x^2)^2 - 34 x^2 y^2 + 25 (y^2)^2

The coefficient of (x^2)^2 is 9 and the coefficient of (y^2)^2 is 25. The product of 9 and 25 is 225. The factors of 225 which sum to -34 are -9 and -25. So 9 (x^2)^2 - 34 x^2 y^2 + 25 (y^2)^2 = 9 (x^2)^2 - 25 x^2 y^2 - 9 x^2 y^2 + 25 (y^2)^2 = 9 x^2 (x^2 - y^2) - 25 y^2 (x^2 - y^2):

9 x^2 (x^2 - y^2) - 25 y^2 (x^2 - y^2)

Factor x^2 - y^2 from 9 x^2 (x^2 - y^2) - 25 y^2 (x^2 - y^2):

(x^2 - y^2) (9 x^2 - 25 y^2)

Factor the difference of two squares. x^2 - y^2 = (x - y) (x + y):

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

9 x^2 - 25 y^2 = (3 x)^2 - (5 y)^2:

(3 x)^2 - (5 y)^2 (x - y) (x + y)

Factor the difference of two squares. (3 x)^2 - (5 y)^2 = (3 x - 5 y) (3 x + 5 y):

Answer: |

| (3 x - 5 y) (3 x + 5 y) (x - y) (x + y)

Answer 2

Aɴꜱᴡᴇʀ

$({x}^{2} - 3 {y}^{2} )(x + 2y)(x - 2y)$

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Sᴛᴇᴘꜱ

${x}^{4} - 7 {x}^{2} {y}^{2} + 22 \times 3 {y}^{4}$

$({x}^{4} - 7 {x}^{2} \times {y}^{2} ) + 22 \times 3 {y}^{4}$

Step -1

Then we shall try to factor a multi variable polynomial

$( {x}^{4} - 7 {x}^{2} {y}^{2} ) + 22 \times 3 {y}^{4}$

factoring the given polynomial using trial and error

Step -2

Factorising

${x}^{2} - 3 {y}^{2}$

we can use the theory- The difference of 2 perfect squares can be factored into

(A+B)(A-B)

Step -3

Factorise

${x}^{2} - 4 {y}^{2}$

Check:4 is the square of 2

${x}^{2}$

is the square of

${x}^{1}$

similarly

${y}^{2}$

is the square of

${y}^{1}$

So factorisation is (x+2y)(x-2y)

Thus the final result

is

$({x}^{2} - 3 {y}^{2} ) (x + 2y)(x - 2y)$

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