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Help!!! will give medal Multiply. x+2/5x-15 times (10x^2-25x-15)
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Choices A. (2x+1)/5(x-3) B. (2x+1)(x-3)/5 C.(10x+5)(x+2) D. (x+2)(2x+1)
well factor the quadratic equation and you get \[5(2x^2 -5x - 3)\] which can be further factored to \[5(x - 3)(2x + 1)\] so this can be written as \[(5x - 15)(2x + 1)\] so the problem is now \[\frac{(x + 2)}{(5x - 15)} \times \frac{(5x - 15)(2x + 1)}{1}\] cancel common factors and you'll have your solution..
what are those @campbell_st
well what do you see that is the same... in the numerator and denominator.... whatever it is... cross it out... what's left is the answer
5x is the same so I cross out tht
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@campbell_st
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