Intermidia Algebra
Andrew factored the expression 12x^328x^23x as 4x (3x^27x+9). But when Melissa applied the distributive law and multiplied out 4x (3x^27x+9), she got 12x^3+28x^236x; thus, Andrew’s solution does not appear to check. Why is that? Please help Andrew to understand this better. Explain your reasoning and correctly factor the original expression, if possible. If the expression is prime, so state.
asked by
Elisa

Given: 12X^3  28X^2  3X.
Andrew made 2 mistakes:
1. You can't factor out 4x, because 3
is not divisible by 4.
2.The 2nd term in the parenthesis should be +7X.
You can only factor out an X:
X(12X^2  28X  3).
posted by Henry
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