Question

find the zeros of the quadratic polynomial p x is equals to x square - 11 x + 28

Answer 1

$\underline{\color{red}hello}$


$x {}^{2} - 11x + 28 = 0 \\ \\ = > x {}^{2} - 4x - 7x + 28 = 0 \\ \\ = > x(x - 4) - 7(x - 4) = 0 \\ \\ = > (x - 4)(x - 7) = 0$


$\therefore \: x - 4 = 0 \\ \\ = > x = 4$


$\therefore \: x - 7 = 0 \\ \\ = > x = 7$


$\colorbox{aqua}{so \: the \: two \: zeroes \: are \: 4 \: and \: 7}$