Find the zeroes of \(3x^2+4x-4\) and verify relations.
\(x=\dfrac{2}{3},\; x=-2\)
Step 1: Split the middle term.
Product = \(3 \times -4 = -12\)
Sum = \(4\)
Numbers = \(6\) and \(-2\)
Step 2: Write and group.
\(3x^2 + 4x - 4 = 3x^2 + 6x - 2x - 4\)
= \((3x^2 + 6x) + (-2x - 4)\)
= \(3x(x + 2) - 2(x + 2)\)
= \((x + 2)(3x - 2)\)
Step 3: Zeroes.
\(x + 2 = 0 \Rightarrow x = -2\)
\(3x - 2 = 0 \Rightarrow x = \dfrac{2}{3}\)
Step 4: Verify.
Sum = \(-2 + \dfrac{2}{3} = -\dfrac{4}{3}\)
= \(-b/a = -4/3\)
Product = \(-2)(\dfrac{2}{3}) = -\dfrac{4}{3}\)
= \(c/a = -4/3\)