Question

5 - (3a^ - 2a) (6 - 3a^ + 2a)

Answer 1

Answer :

Now,

5 - (3a² - 2a) (6 - 3a² + 2a)

= 5 - x (6 - x), let 3a² - 2a = x

= 5 - 6x + x²

= x² - 6x + 5

= x² - x - 5x + 5

= x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

= (3a² - 2a - 1) (3a² - 2a - 5),
since x = 3a² - 2a

= (3a² - 3a + a - 1) (3a² + 3a - 5a - 5)

= {3a (a - 1) + 1 (a - 1)} {3a (a + 1) - 5 (a + 1)}

= (a - 1) (3a + 1) (a + 1) (3a - 5),

which is the required factorization.

#MarkAsBrainliest

Answer 2

Your question needs a correction,

Correct question, 5-(3a²-2a)(6-3a²+2a)

××××××××××××××××××××××××

5 - (3a² - 2a) [ 6 - (3a² - 2a)]

Let, 3a² - 2a = x

5 - x(6 - x)

5 - 6x + x²

5 - (5 + 1)x + x²

5 - 5x - x + x²

5(1 - x) - x(1 - x)

(1 - x) (5 - x)

(1 - 3a² + 2a) (5 - 3a² + 2a)

( -3a² + 2a + 1)(-3a² + 2a + 5)

(-3a² + {3 - 1}a + 1) (-3a² + {5 - 3}a + 5)

( -3a² + 3a - a + 1)(-3a² + 5a - 3a + 5)

[ -3a(a - 1)-1(a - 1)] [ -a(3a -5) -1(3a - 5)]

(-3a - 1)(a - 1)(3a - 5)(-a - 1)


I hope this help you


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