Question

factorise
15(2x-3)^3-10(2x-3)​

Answer 1

Step by Step Solution:

STEP

1

:

Equation at the end of step 1

(15 • ((2x - 3)3)) - 10 • (2x - 3)

STEP

2

:

Equation at the end of step 2

15 • (2x - 3)3 - 10 • (2x - 3)

STEP

3

:

Pulling out like terms

3.1 Pull out 2x-3

After pulling out, we are left with :

(2x-3) • ( 15 * (2x-3)2 +( 10 * (-1) )) 3.2 Evaluate : (2x-3)2 = 4x2-12x+9

STEP

4

:

Pulling out like terms

4.1 Pull out like factors :

60x2 - 180x + 125 = 5 • (12x2 - 36x + 25)

Trying to factor by splitting the middle term

4.2 Factoring 12x2 - 36x + 25

The first term is, 12x2 its coefficient is 12 .

The middle term is, -36x its coefficient is -36 .

The last term, "the constant", is +25

Step-1 : Multiply the coefficient of the first term by the constant 12 • 25 = 300

Step-2 : Find two factors of 300 whose sum equals the coefficient of the middle term, which is -36 .

-300 + -1 = -301

-150 + -2 = -152

-100 + -3 = -103

-75 + -4 = -79

-60 + -5 = -65

-50 + -6 = -56

For tidiness, printing of 30 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Final result :

5 • (2x - 3) • (12x2 - 36x + 25)

Answer 2

Answer:

Step-by-step explanation:  

Take 5( 2 x - 3 )  common    

5 ( 2 x - 3 ) ( 3 ( 2 x - 3 )^2  - 2)

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

5 ( 2x - 3 ) ( 12x^2 - 12x - 9 -2 )

                                          ^^^^  =   -11    

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