You can solve this problem through factoring.
First, you have the equation,
Then, you can factor the numerator.
You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.
You are left with 
The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.
one would say that the simple interest doubles if the period of time is specified in the contract and the contract is still valid, if the interest amount is available anitime and so on.
So if the amount doubles let's say at half time for which the principal was awarded to the bank, by the end of the contract , the interest amount can be double × just increased by 1.5
Answer:
<h3>
f(x) = 5x² + 2x</h3><h3>
g(x) = 6x - 6</h3>
Step-by-step explanation:
![\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}\\\\6(x^2-3x+2)\ne0\ \iff\ x=\frac{3\pm\sqrt{9-8}}{2}\ne0\ \iff\ x\ne2\ \wedge\ x\ne1\\\\\\\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}=\dfrac{x(5x^2-8x-4)}{6(x^2-3x+2)}=\dfrac{x(5x^2-10x+2x-4)}{6(x^2-2x-x+2)}=\\\\\\=\dfrac{x[5x(x-2)+2(x-2)]}{6[x(x-2)-(x-2)]} =\dfrac{x(x-2)(5x+2)}{6(x-2)(x-1)}=\dfrac{x(5x+2)}{6(x-1)}=\dfrac{5x^2+2x}{6x-6}\\\\\\f(x)=5x^2+2x\\\\g(x)=6x-6](https://tex.z-dn.net/?f=%5Cdfrac%7B5x%5E3-8x%5E2-4x%7D%7B6x%5E2-18x%2B12%7D%5C%5C%5C%5C6%28x%5E2-3x%2B2%29%5Cne0%5C%20%5Ciff%5C%20x%3D%5Cfrac%7B3%5Cpm%5Csqrt%7B9-8%7D%7D%7B2%7D%5Cne0%5C%20%5Ciff%5C%20x%5Cne2%5C%20%5Cwedge%5C%20x%5Cne1%5C%5C%5C%5C%5C%5C%5Cdfrac%7B5x%5E3-8x%5E2-4x%7D%7B6x%5E2-18x%2B12%7D%3D%5Cdfrac%7Bx%285x%5E2-8x-4%29%7D%7B6%28x%5E2-3x%2B2%29%7D%3D%5Cdfrac%7Bx%285x%5E2-10x%2B2x-4%29%7D%7B6%28x%5E2-2x-x%2B2%29%7D%3D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7Bx%5B5x%28x-2%29%2B2%28x-2%29%5D%7D%7B6%5Bx%28x-2%29-%28x-2%29%5D%7D%20%3D%5Cdfrac%7Bx%28x-2%29%285x%2B2%29%7D%7B6%28x-2%29%28x-1%29%7D%3D%5Cdfrac%7Bx%285x%2B2%29%7D%7B6%28x-1%29%7D%3D%5Cdfrac%7B5x%5E2%2B2x%7D%7B6x-6%7D%5C%5C%5C%5C%5C%5Cf%28x%29%3D5x%5E2%2B2x%5C%5C%5C%5Cg%28x%29%3D6x-6)
Answer:
(x - 5) (x (4 x - 3) + 8)
Step-by-step explanation:
4 x^3 - 23 x^2 + 23 x - 40
x (x (4 x - 23) + 23) - 40
(x - 5) (x (4 x - 3) + 8)
