Question

Please help me Factorize these questions ​

Answer 1

13. a^3 + 27

a^3 + 3^3

(a+3)(a^2 -3a +9)

14. b^3 -125

b^3 - 5^3

(b - 3)(b^2 +2b + 25

15. -5k^2 - 5k + 280

-5(k^2 + k - 56)

-5(k^2 + 8k - 7k - 56)

-5(k+ 8)(k- 7)

16. 3m^2 + 63mn - 300n^2

3(m^2 + 21mn - 100n^2)

3(m + 25n)(m - 4n)

17. 3x^3 - 24y^3

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

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

18. ax^2 - bx^2 - 4ay^2 + 4by^2

x^2(a - b) - 4y^2(a + b)

(x^2 - 4y^2)(a^2 - b^2)

(x - 2y)(x + 2y)(a + b)(a - b)

Answer 2

Answer:

13. (a + 3)(a² - 3a + 9)

14. (b - 5)(b² + 5b + 25)

15. -5 (k + 8)(k - 7)

16. 3(m + 25n)(m - 4n)

17. 3(x - 2y)(x² + 2xy + 4y²)

18. (a - b)(x - 2y)(x + 2y)

Step-by-step explanation:

Given :

13. a³ + 27

14. b³ - 125

15. -5k² - 5k + 280

16. 3m² + 63mn - 300n²

17. 3x³ - 24y³

18. ax² - bx² - 4ay² + 4by²

To find :

Factorize the given equations

Formulas Used:

$\boxed{\mathrm{p^3+q^3 = (p+q)(p^2-pq+q^2)}}$

$\boxed{\mathrm{p^3-q^3 = (p-q)(p^2+pq+q^2)}}$

Solution :

13. a³ + 27

=> Here, p = a and q = 3

So,

a³ + 27 = (a + 3)(a² - 3a + 9)

__________________

14. b³ - 125

=> Here, p = b and q = 5

So,

b³ - 125 = (b - 5)(b² + 5b + 25)

_____________________

15. -5k² - 5k + 280

=> -5 (k² + k - 56)

=> -5 (k² + 8k - 7k - 56)

=> -5 (k(k + 8) -7 (k + 8))

=> -5 (k + 8)(k - 7)

____________________

16. 3m² + 63mn - 300n²

=> 3(m² + 21mn - 100n²)

=> 3(m² + 25mn - 4mn + 100n²)

=> 3(m(m + 25n) - 4n(m + 25n))

=> 3(m + 25n)(m - 4n)

______________________

17. 3x³ - 24y³

=> 3 (x³ - 8y³)

Here, p = x and q = 2y

So,

3(x³ - 8y³) = 3(x - 2y)(x² + 2xy + 4y²)

__________________________

18. ax² - bx² - 4ay² + 4by²

=> x²(a - b) - 4y²(a - b)

=> (a - b)(x² - 4y²)

We know that,

$\boxed{\mathrm{x^{2} -y^{2} = (x+y)(x-y)}}$

so,

=> (a - b)(x - 2y)(x + 2y)

Hope it helps!!