Question

Please factor and find the zeros of each relation (18 marks)

Answer 1

Answer:

$C. \ \frac{-2}{3} , \ \frac{-4}{5}\\\\D. \ -2, \ -4$

Step-by-step explanation:

C.

$30x^2 + 44x +16 = 0\\\\2(15x^2 + 22 + 8) = 0\\\\2(15x^2 + 12x + 10x + 8) = 0\\\\2(3x(5x + 4)+2(5x + 4))=0\\\\2((3x + 2)(5x + 4)) = 0\\\\Zeroes : 3x + 2 = 0 => x = \frac{-2}{3}$

             $5x+ 4 = 0 => x = \frac{-4}{5}$

D.

$10x^2 + 60x +80 = 0\\\\10(x^2 + 6x + 8)= 0\\\\10(x^2 + 4x + 2x + 8)=0\\\\10(x(x+4)+2(x+4))=0\\\\10((x+2)(x+4))=0\\\\Zeroes : x + 2 = 0 =>x = -2$

            $x + 4 = 0 => x= -4$