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