To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
The answer is
<span>y = four thirdsx + 8
</span>
Answer:
Step-by-step explanation:
x²+5x+6=x²+3x+2x+6=x(x+3)+2(x+3)=(x+3)(x+2)
so Volume=(x+3)(x+2)(x+4)
V=lwh
one of the factors is the length.
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)
