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What is the completely factored form of this polynomial: (18x^3 - 120x^2 - 40x)?

A. (6x(3x + 1)(x - 7))
B. (6x(3x - 1)(x + 7))
C. (6x(3x - 1)(x - 7))
D. (6x(3x + 1)(x + 7))

User JulienD
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1 Answer

7 votes

Final answer:

The polynomial 18x^3 - 120x^2 - 40x can be factorized by first taking out the common factor 6x, and then factoring the remaining quadratic. The factored form is 6x(5)(3x - 1)(x - 7), which is option C.

Step-by-step explanation:

The question asks for the completely factored form of the polynomial 18x^3 - 120x^2 - 40x. To factor this polynomial, we will first look for a common factor in all terms. We can see that 6x can be factored out, which gives us:



6x(3x^2 - 20x - 20)



Now, we need to factor the quadratic expression 3x^2 - 20x - 20. By looking for two numbers that multiply to (3)(-20) = -60 and add to -20, we find that -25 and 5 fulfill this condition. So, we rewrite the middle term -20x as -25x + 5x and factor by grouping.



6x(3x^2 - 25x + 5x - 20)



Grouping the terms, we get:



6x((3x^2 - 25x) + (5x - 20))



Factor further:



6x(x(3x - 25) + 5(3x - 1))



Since (3x - 25) and (3x - 1) are not the same, we cannot factor them as a common factor. However, we can divide 25 by 5, which gives us:



6x(5x(3x - 5) + 5(3x - 1))



Now it is clear that we have a common factor of 5:



6x * 5(x(3x - 5) + (3x - 1))



Which simplifies to:



6x(5)(3x - 1)(x - 7), This is the factored form, which corresponds to option C.

User Katayoun
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