Answer:
Try to search up euphemisms to help you in an argument where your forced to say no
Step-by-step explanation: In this situation apologize to them and say
I can't do that or no
Answer:
is the required factorization of f(x).
Step-by-step explanation:
To factor the expression we must first group the terms and then take out common from these groups
![f(x)=x^3+3x^2+16x+48=(x^3+3x^2)+(16x+48)](https://tex.z-dn.net/?f=%20f%28x%29%3Dx%5E3%2B3x%5E2%2B16x%2B48%3D%28x%5E3%2B3x%5E2%29%2B%2816x%2B48%29)
Taking
common from first group and the 16 from second group we get:
![f(x) = x^2(x+3)+16(x+3) = (x+3)(x^2+16)](https://tex.z-dn.net/?f=%20f%28x%29%20%3D%20x%5E2%28x%2B3%29%2B16%28x%2B3%29%20%3D%20%28x%2B3%29%28x%5E2%2B16%29%20)
Now, to factor in complex from we have to break term ![x^2+16](https://tex.z-dn.net/?f=%20x%5E2%2B16)
![f(x)= (x+3){x^2-(-4i)^2}](https://tex.z-dn.net/?f=%20f%28x%29%3D%20%28x%2B3%29%7Bx%5E2-%28-4i%29%5E2%7D%20)
As, ![i^2 = -1 , therefore (-4i)^2 = 16i^2 =-16](https://tex.z-dn.net/?f=%20i%5E2%20%3D%20-1%20%2C%20therefore%20%28-4i%29%5E2%20%3D%2016i%5E2%20%3D-16%20)
Also using identity ![a^2-b^2 =(a+b)(a-b)](https://tex.z-dn.net/?f=%20a%5E2-b%5E2%20%3D%28a%2Bb%29%28a-b%29%20)
On solving
![f(x) = (x+3)(x+4i)(x-4i)](https://tex.z-dn.net/?f=%20f%28x%29%20%3D%20%28x%2B3%29%28x%2B4i%29%28x-4i%29%20)
is the required factorization of f(x).
Simplify the expression
-10
Hope this helps! :)