You should use quadratic factorization to write the numerator and denominator of the fraction `(x^2-3x-10)/(x^2+7x+10).`

You need to find the roots of `x^2-3x-10` . Use quadratic formula:

`x_(1,2) = (3+-sqrt(9+40))/2 =gt x_(1,2) = (3+-sqrt(49))/2`

`=gt x_(1,2) = (3+-7)/2 =gt x_1=(3+7)/2=5`

`=gt x_2 = (3-7)/2 = -2`

Using quadratic factorization yields: `x^2-3x-10 = (x-x_1)(x-x_2) =gt x^2-3x-10 = (x-5)(x+2)`

`` You need to find the roots of `x^2+7x+10` . Use quadratic formula:

`x_(1,2) = (-7+-sqrt(49-40))/2 =gt x_(1,2) = (-7+-sqrt(9))/2`

`=gt x_(1,2) = (-7+-3)/2 =gt x_1=(-7+3)/2=-2`

`=gt x_2 = (-3-7)/2 = -5`

Using quadratic factorization yields: `x^2+7x+10 = (x-x_1)(x-x_2) =gt x^2+7x+10 = (x+5)(x+2)`

Using quadratic factorization yields: `(x^2-3x-10)/(x^2+7x+10) = ((x-5)(x+2))/((x+5)(x+2)) = (x-5)/(x+5)`

**Equating (x-5)/(x+5) = (x-a)/(x+b) yields a = 5 and b = 5.**

first what is the mutiple of -10 it's either -5*2 or -2*5 or-10*1 or-1*10

then how u get -3 .With 5 and 2 u can get -3 it's -5 +2 therefore x^2-3x-10=(x-5)(x+2)

x^2+7x+10

the mutiple of the +10 is either 5*2 or 1*10 but what will give u 7 it's 5+2 therefore x^2+7x+10=(x+2)(x+5)

(x^2-3x-10)/(x^2+7x+10)=((x-5)(x+2))/((x+2)(x+5))

=(x-5)/(x+5)

a=5 and b= 5