Question

find the zeroes of the linear and quadratic polynomial of x^2-3x-4. answer in method​

Answer 1

${x}^{2} - 3x - 4 = 0 \\ = > {x}^{2} - 4x + x - 4 = 0 \\ = > x(x - 4) + 1(x - 4) = 0 \\ = > (x - 4)(x + 1) = 0 \\ \\ so..... \\ x - 4 = 0 \: \: \: \: or \: x + 1 = 0 \\ = > x = 4 \: \: \: \: \: or \: \: \: x = - 1$

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