Question

Tutorial ExerciseMultiply.x²yAx2 + 2x - 8.хбу?x2 + 11x + 28Click here to begin!Need Help?Watch ItAdditional MaterialseBook

Answer 1

From the given problem :

`(x^2y^4)/(x^2+11x+28)*(x^2+2x-8)/(x^6y^3)`

x^2 + 11x + 28 can be factored as, you need to think of two numbers that has a product of 28 and a sum of 11. In this case it's 7 and 4 :

(x + 7)(x + 4)

and x^2 + 2x - 8 can be factored as, you need to think of two numbers that has a product of -8 and a sum of 2. In this case it's 4 and -2 :

(x + 4)(x - 2)

So the expression will be :

`(x^2y^4)/((x+7)(x+4))*((x+4)(x-2))/(x^6y^3)`

The term (x+4) will be cancelled out

`(x^2y^4(x-2))/(x^6y^3(x+7))`

Simplify further and the answer is :

`(y(x-2))/(x^4(x+7))`