Question

write the second degree polynomial x²-16 as the product of two first degree polynomials.​

Answer 1

Step-by-step explanation:

So, a quadratic equation is

$a {x}^{2} + bx + c$

where, -b/a= sum of the roots

and c/a = product of roots

let α, β be the roots of this equation then

$(x - \alpha )(x - \beta )$

is the quadratic equation

now here your question is,

${x}^{2} - 16$

first method: compare this with

ax^2+bx+c

so here b=0 and a=1, c = -16.

α+β= -b/a= 0 ( since b=0 in your question)

from this equation we get α= -β.

αβ=c/a =-16/1 = -16.

we found α= -β above so put α= -β in above equation

αβ= -16,

$- { \alpha }^{2} = - 16 \\ { \alpha }^{2} = 16 \\ \alpha = 4 \: or \: - 4$

then β= -4 if α= 4 and β= 4 if α= -4.

put these value in

$(x - \alpha )(x - \beta )$

and your answer will be

$(x + 4)(x - 4)$

Alternate method:

just use formula

${a}^{2} - {b}^{2} = (a + b)(a - b)$

so

${x}^{2} - 16 \\ = {x}^{2} - {4}^{2} \\ = (x + 4)(x - 4)$

HOPE IT HELPS