SOLUTION: Andrew factored the expression28x^2-42x^2+35x as 7x(4x^2+6x-5) . But when Melissa applied the distributive law and multiplied out7x(4x^2+6x-5) , she got28x^3+24x^2-23x ; thus, Andr

Algebra ->  Square-cubic-other-roots -> SOLUTION: Andrew factored the expression28x^2-42x^2+35x as 7x(4x^2+6x-5) . But when Melissa applied the distributive law and multiplied out7x(4x^2+6x-5) , she got28x^3+24x^2-23x ; thus, Andr      Log On


   

Question 489000: Andrew factored the expression28x^2-42x^2+35x as 7x(4x^2+6x-5) . But when Melissa applied the distributive law and multiplied out7x(4x^2+6x-5) , she got28x^3+24x^2-23x ; 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
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

Andrew factored the expression
28x%5E2-42x%5E2%2B35x as 7x%284x%5E2%2B6x-5%29 .
But when Melissa applied the distributive law and multiplied out7x(4x^2+6x-5) , she got
28x%5E3%2B24x%5E2-23x+;

let's check first Melisa's result:
7x%284x%5E2%2B6x-5%29
7x%2A4x%5E2%2B6x%2A7x-5%2A7x
28%5E3%2B42x%5E2-35x....she made these mistakes when multiplied 7x%284x%5E2%2B6x-5%29: for 6x%2A7x-5%2A7x she got 24x%5E2-23x+...(6%2A7=42 and -5%2A7=-35


but, Andrew’s solution does not check either because he made mistake with a minus sign
in expression 28x%5E2-42x%5E2%2B35x coefficients are 28 (positive), -42 (negative), and 35 (positive)
so, instead of 7x%284x%5E2%2B6x-5%29, he should have 7x%284x%5E2-6x%2B5%29

I hope Andrew will understand this better now. So, be careful with the signs and do a quick check that the distributive law takes you back to the expression you started with.
correctly factored original expression: 7x%284x%5E2-6x%2B5%29

If the expression is prime: It isn't prime if it have factors besides 1 and itself.