Question

guys plz do answer me for the 4th question.... irrelevant answers will be reported on the spot....​

Answer 1

Answer:

(x-9)(x-12)

its factors are given above

factorization numbers are 9 and 12.

thanks

hope u understood

pls mark brainliest

if u need more help pls ask.

Answer 2

Method 1 ⇒ Factorization

Let's solve your equation step-by-step.

$x^{2}-21x+108=0$

Step 1: Factor left side of equation.

⇒ $(x-9)(x-12)=0$

Step 2: Set factors equal to 0.

⇒ $x-9=0 \ or \ x-12=0$

⇒ $x=9 \ or \ x=12$

$\rule{300}{1}$

Method 2 ⇒ With Quadratic Formula

Let's solve your equation step-by-step.

$x^{2}-21x+108=0$

For this equation: a = 1, b = -21, c = 108

⇒ $1x^{2}+-21x+108=0$

Step 1: Use quadratic formula with a = 1, b = -21, c = 108.

⇒ $x = \dfrac{-b\± \sqrt{b^{2} - 4ac}}{2a}$

⇒ $x = \dfrac{-(-21)\± \sqrt{(-21)^{2} - 4(1)(108)}}{2(1)}$

⇒ $x = \dfrac{21\± \sqrt{441 - 4(108)}}{2}$

⇒ $x = \dfrac{21\± \sqrt{441 - 432}}{2}$

⇒ $x = \dfrac{21\± \sqrt{9}}{2}$

⇒ $x = \dfrac{21\± 3}{2}$

⇒ $x=12 \ or \ x=9$

$\rule{300}{1}$

$\bf \therefore x=9 \ or \ x=12 \ in \ the \ equation => \ x^{2}-21x+108=0$

$\rule{300}{1}$