Pre-calculus Factoring Quadratic Equations: 0.4x^2 + 0.6x - 1.8

VixxenBlack said:
Hi, I really need help to factor this decimal question please!

0.4x^2 + 0.6x - 1.8
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Get rid of the decimals by multiplying by 10. BUT! Note that this isn't an equation (there is no "equals" sign), so there's nothing to "multiply through". Instead, you have to multiply AND DIVIDE by 10!

. . . . .\(\displaystyle \left(\dfrac{1}{10}\right)\, (10)\, (0.4\, x^2\, +\, 0.6\, x\, -\, 1.8)\)

Multiply just the 10 inside, leaving the 1/10 outside:

. . . . .\(\displaystyle \left(\dfrac{1}{10}\right)\, (4\, x^2\, +\, 6\, x\, -\, 18)\)

Note now that you can divide out a 2:

. . . . .\(\displaystyle \left(\dfrac{1}{10}\right)\, (2)\, (2x^2\, +\, 3x\, -\, 9)\)

Now factor inside. When you're done inside, you can multiply the 1/10 and the 2 back inside, or not. ;)
 
Multiplying the original equation by 5 will eliminate the decimal fractions more simply:

. . .\(\displaystyle 0.4x^2\,+\, 0.6x\,+\, 0.8\,= \,\dfrac{1}{5}\,(2x^2\,+\, 3x\,+\, 4)\).

Of course that is exactly the kind of thing I would not see at the start but once you have \(\displaystyle \frac{2}{10}\) it is kind of obvious!
 
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