SOLUTION: can i please get help with this? x^2+2x-15 over 4x^2 divided by x^2-25 over 2x-10

Click here to see ALL problems on Polynomials-and-rational-expressions Question 77368: can i please get help with this? x^2+2x-15 over 4x^2 divided by x^2-25 over 2x-10 Found 2 solutions by jim_thompson5910, stanbon: Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website! Factor the top most numerator Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1) In order to factor , first we need to ask ourselves: What two numbers multiply to -15 and add to 2? Lets find out by listing all of the possible factors of -15 Factors: 1,3,5,15, -1,-3,-5,-15,List the negative factors as well. This will allow us to find all possible combinations These factors pair up to multiply to -15. (-1)*(15)=-15 (-3)*(5)=-15 Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2 |||| First Number | Second Number | Sum 1 | -15 | 1+(-15)=-14 3 | -5 | 3+(-5)=-2 -1 | 15 | (-1)+15=14 -3 | 5 | (-3)+5=2 We can see from the table that -3 and 5 add to 2.So the two numbers that multiply to -15 and add to 2 are: -3 and 5 Now we substitute these numbers into a and b of the general equation of a product of linear factors which is: substitute a=-3 and b=5 So the equation becomes: (x-3)(x+5) Notice that if we foil (x-3)(x+5) we get the quadratic again Factor the bottom numerator Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1) In order to factor , first we need to ask ourselves: What two numbers multiply to -25 and add to 0? Lets find out by listing all of the possible factors of -25 Factors: 1,5,25,15, -1,-5,-25,-15,List the negative factors as well. This will allow us to find all possible combinations These factors pair up to multiply to -25. (-1)*(15)=-25 (-5)*(25)=-25 Now which of these pairs add to 0? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 0 |||| First Number | Second Number | Sum 1 | -15 | 1+(-15)=-14 5 | -25 | 5+(-25)=-20 -1 | 15 | (-1)+15=14 -5 | 25 | (-5)+25=20 substitute a=-3 and b=5 So the equation becomes: (x-3)(x+5) Notice that if we foil (x-3)(x+5) we get the quadratic again None of these factors add to 0. So this quadratic cannot be factored. In order to solve for x, we need to use the quadratic formula. So now we have Factor a 2 out of Flip the 2nd fraction and multiply Cancel like terms Reduce. This is the simplified form. Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website! x^2+2x-15 over 4x^2 divided by x^2-25 over 2x-10 ----------- Factor where you can to get: (x+5)(x-3) over 4x^2 divided by (x+5)(x-5) over 2(x-5) -------------- Invert the denominator and change to multiplication. (x+5)(x-3) over 4x^2 muliplied by 2(x-5) over (x=5)(x+5) ------------- Cancel any factors common to a numerator and a denominator to get: (x-3) over 2x^2 ============ Cheers, Stan H.