Beginning Algebra
14x2+3x+9\(\displaystyle\) \(\frac{1}{4}\)x^2 + 3x + 9
12x2+3x+9\(\displaystyle\) \(\frac{1}{2}\)x^2 + 3x + 9
14x2+32x+9\(\displaystyle\) \(\frac{1}{4}\)x^2 + \(\frac{3}{2}\)x + 9
14x2+6x+9\(\displaystyle\]\frac{1}{4}\)x^2+6x+9