Question

fund the zeroes p(x)=x^2-6x+9​

Answer 1

Answer:

x^2-3x-3x+9

x(x-3)-3(x-3)

(x-3)(x-3)

Answer 2

QUESTION:

find the zeroes p(x)=x^2-6x+9

ANSWER:

Quadratic equation is in the form of

$a {x}^{2} + bx + c = 0$

we have to factorised it to find the zeroes.

We split the middle term in such a way that it's sum equal to b i.e -6 and product equal to a ×c that is 9×1.

${x}^{2} - 6x + 9 \\ \\ {x}^{2} - 3x - 3x + 9 \\ x(x - 3) - 3(x - 3) \\ (x - 3)(x - 3)$

now;

$x - 3 = 0 \\ x = 3$

FINAL ANSWER :

zeroes are 3.