Question

-3x(7x-4)+6x-(13-24x²)​

Answer 1

Answer:

• $3x^2+18x-13.$

Step-by-step 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

Answer:

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

Step-by-step 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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