Recent content by BestMethod

$$(x^2-5x+16)/(2x+1)(x-2)^2=a/(2x+1) + b/(x-2) + c/(x-2)^2$$ To find a,multiply both sides by $$(2x+1)$$; $$(x^2-5x+16)/(x-2)^2=a + b(2x+1)/(x-2) + c(2x+1)/(x-2)^2$$ Now,set 2x+1=0 or x=-1/2, 3=a+0+0=a To find c,multiply both sides by $$(x-2)^2$$; $$(x^2-5x+16)/(2x+1)=a(x-2)^2/(2x+1) +...

Let's factor y=(2x^2 +13x-7) this way; If we multiply & divide the polynomial(let’s call it f) by a number,say m,f won’t be changed; mf/m=(m/m)f=1*f=f Now,one way here is to multiply & divide f by 2; 2(2x^2 +13x-7)/2 = (4x^2 +26x-14)/2 = [(2x)^2+13(2x)-14]/2 Taking 2x=z; y=(z^2+13z-14)/2...

(z^2-4z+4)/(z-3)(z-1)^2 = 1/(z-3) - 2/(z-1)^2 + C/(z-1) Now,plugging any number into z,say 0; -4/3 = -1/3 -2 -C .Then, C=-1

Let's use a better method; (5x^2+1)/(3x+2)(x^2+3) = A/(3x+2) + (Bx+C)/(x^2+3) To find A,multiplying both sides of the equality by (3x+2) ; (5x^2+1)/(x^2+3) =A + (3x+2)(Bx+C)/(x^2+3) (1) Now,setting x=-2/3 or 3x+2=0,makes the expression on the right containing (3x+2)...

Now,to find a,multiplying both sides of the equality by (x+3); (x-3)/(x+1) =A + B(x+3)/(x+1) Now,setting x=-3 or x+3=0,makes the expression on the right containing (x+3) vanish. So, (-3-3)/(-3+1)=A+0 .Then, A=3 You could find B by a similar method.

In this kind of factoring,a good way is the following; As a result of the remainder theorem,if x=b makes any polynomial vanish,then (x-b) is a factor of it. Now,how could you find any number such b? According to the rational root theorem,a possible integer root of such a polynomial should be...

To find A,multiplying the both sides by (x-2); x^2/(x+3)(x-1)=A + B(x-2)/(x+3) + C(x-2)/(x-1) Now,by setting x=2,the expressions containing (x-2) vanish,so we get rid of B & C, 2^2/(2+3)(2-1)=A+0+0, then, 4/5=A By the similar method we could find B & C.

(6x-5)/(x-4)(x²+3)= a/(x-4) + (bx+c)/(x²+3) Now,to find a,multiplying both sides of the equality by (x-4); (6x-5)/(x²+3) = a + (x-4)(bx+c)/(x²+3). (1) Now,setting x=4 or x-4=0,makes the expression on the right vanish,so (24-5)/(16+4)=a+0 a=19/20 To find c,set x=0 in (1),so we get rid of...

As a result of the remainder theorem,if x=b makes any polynomial vanish,then (x-b) is a factor of it. Now,how could you find any number such b? According to the rational root theorem,a possible integer root of such an equation should be a divisor of the constant term(here is 4) which are ...