Question

Factor out the largest possible term.

Answer 1

Answer: 4x^2*(3x^3 - 1 + 4x)

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Work Shown:

The GCF of the coefficients 12,-4,16 is 4
The GCF of the variable terms is x^2 since this is the smallest variable term that is found in each of x^5, x^2 and x^3
Overall, the GCF is 4x^2. 

Factor each term in terms of the GCF
12x^5 = (12)*(x^5)
12x^5 = (4*3)*(x^3*x^2)
12x^5 = (4x^2)*(3x^3) <--- call this equation (1)
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4x^2 = (4x^2)*(1) <--- call this equation (2)
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16x^3 = (4*4)*(x^2*x)
16x^3 = (4x^2)*(4x) <--- call this equation (3)
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Using those individual factorizations, we can apply the distributive property to factor out the GCF

12x^5 - 4x^2 + 16x^3
(4x^2)*(3x^3) - 4x^2 + 16x^3 ... substitute equation (1)
(4x^2)*(3x^3) - (4x^2)*(1) + 16x^3 ... substitute equation (2)
(4x^2)*(3x^3) - (4x^2)*(1) + (4x^2)*(4x) ... substitute equation (3)
4x^2*(3x^3 - 1 + 4x) ... use the distributive property

Therefore, 12x^5 - 4x^2 + 16x^3 factors to 4x^2*(3x^3 - 1 + 4x)

This means, 12x^5 - 4x^2 + 16x^3 = 4x^2*(3x^3 - 1 + 4x) is a true equation for all real numbers x.