Question

Solve this eqution. 5x^3+20x^2-105x/5x^2-15x=0

Answer 1

We have
\[\frac{5x^3+20x^2-105x}{5x^2-15x}=0\]
Let's Take 5x common from both numerator and denominator
\[\frac{5x(x^2+4x-21)}{5x(x-3)}=0\]
Cancel the 5x from both numerator and denominator
\[\frac{\cancel {5x}(x^2+4x-21)}{\cancel {5x}(x-3)}=0\]
so we have now
\[\frac{x^2+4x-21}{x-3}=0\]
Now let's find the factors of
\[x^2+4x-21\]
Compare this with
\[ax^2+bx+c\]
a=1
b=4
c=-21
Let's find the factors of ab=-21 such that the sum of the factors is 4 (b)
7 and -3 satisfy this
7 *-3=-21
(7)+(-3)=4
so
\[4x=7x-3x\]
we have now
\[x^2+7x-3x-21\]
take x common from the first two terms and -3 from last two terms
\[x(x+7)-3(x+7)\]
Take (x+7) common now
\[(x-3)(x+7)\]
So we have now
\[\frac{(x-3)(x+7)}{x-3}=0\]
Now cancel x-3 from both numerator and denominator
\[\frac{\cancel{(x-3)}(x+7)}{\cancel{x-3}}=0\]
we have now
\[x+7=0\]
so we get
\[x=-7\]