Question

Determine the value of the q and p x^{2} -qx-28=(x+7)(x-p)

Answer 1

Answer:

rhs

= (x+7)(x-p)

$= x(x - p) + 7(x - p)$

$= {x}^{2} - px + 7x - 7p$

$= {x}^{2} + ( - p + 7)x - 7p$

lhs = rhs

${x}^{2} - qx - 28 = {x}^{2} + ( - p + 7)x - 7p$

-7p

$- qx - 28 = ( - p + 7)x - 7p$

$therefore. \: \: \: - qx = ( - p + 7)x$

$and \: - 28 = - 7p$

p=4

$- qx = ( - 4 + 7)x$

$- q = 3$

$q = - 3$