Question

-3x(7x-4)+6x-(13-24x²)​

Answer 1

• 
  `3x^2+18x-13.`

Explanation:

Step 1:

Equation should be:

• 
  `((0-(3x*(7x-4)))+6)-(13-(2^3*3x^2))`

Step 2:

Equation should be:

• 
  `((0-3x*(7x-4))+6x)-(13-24x^2)`

Step 3:

• Factoring
  `3x^2+18x-13`
• The first term is
  `3x^2`. It's coefficient is 3.
• The middle term is +18x, it's coefficient is 18.
• The last term, "the constant", is -13.

Step 1:

• Multiply the coefficient of the first term by the constant 3 * -13 = -39

Step 2:

• Find two factors of -39 whose sum equals the coefficient of the middle term, which is 18 .
• -39 + 1 = -38
• -13 + 3 = -10
• -3 + 13 = 10
• -1 + 39 = 38
• Observation : No two such factors can be found !!
• Conclusion : Trinomial can not be factored

Solution:

• 
  `3x^2 + 18x - 13`.

Answer 2

`\boxed{\tt \tt 3x^2+18x-13}`

Explanation:

`\tt -3x(7x-4)+6x-(13-24x^2)`

Apply the Distributive Property

`\tt -3x\left(7x-4\right)+6x-13+24x^2`

Multiply, -3 by (7x-4):

`\tt -21x^2+12x+6x-13+24x^2`

Combine like terms:

`\tt (-21x^2+24x^2)+(12x+6x)-13`

`\tt 3x^2+18x-13`

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