SOLUTION: Find the error in each of the following and provide an explanation of the mistake made. a). 𝑥2+25= (𝑥+5)(𝑥+5) b). 10𝑥3+15𝑥2+5&#1

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find the error in each of the following and provide an explanation of the mistake made. a). 𝑥2+25= (𝑥+5)(𝑥+5) b). 10𝑥3+15𝑥2+5&#1      Log On


   

Question 999760: Find the error in each of the following and provide an explanation of the mistake made.
a). 𝑥2+25= (𝑥+5)(𝑥+5)
b). 10𝑥3+15𝑥2+5𝑥 = 5𝑥(2𝑥2+3𝑥)



c). 𝑥2−2𝑥+24= (𝑥−6)(𝑥+4)

Found 2 solutions by ankor@dixie-net.com, MathLover1:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Find the error in each of the following and provide an explanation of the mistake made.
a). 𝑥^2 + 25 = (𝑥+5)(𝑥+5)
If you FOIL the two factors you have you get: x^2+10x+25,
x^2 + 25 does not have any real roots
:
b). 10𝑥^3+15𝑥^2+5𝑥 = 5𝑥(2𝑥^2+3𝑥) it should be 5x(2x^2 + 3x + 1)
:
c). 𝑥^2 − 2𝑥 + 24 = (𝑥−6)(𝑥+4) if you FOIL this you get x^2 - 2x - 24 (-6*4)
equation does not have any real roots

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

a).
x%5E2%2B25=+%28x%2B5%29%28x%2B5%29 => this is not equal,
x%5E2%2B5%5E2%3C%3E+%28x%2B5%29%28x%2B5%29 because x%5E2%2B5%5E2 is the sum of squares and the rule for factoring the sum of squares:a%5E2%2Bb%5E2 can be factorized as %28a%2Bib%29%28a-ib%29
if your factors were %28x-5i%29%28x%2B5i%29, it would work because x%5E2%2B25=%28x-5i%29%28x%2B5i%29

b).
10x%5E3%2B15x%5E2%2B5x+=+5x%282x%5E2%2B3x%29+....missing one more term in parentheses
10x%5E3%2B15x%5E2%2B5x+=+5x%282x%5E2%2B3x%2B1%29+



c).
x%5E2-2x%2B24=+%28x-6%29%28x%2B4%29+=> multiply %28x-6%29%28x%2B4%29+=>x%5E2%2B4x-6x-24=x%5E2-2x-24+; as you can see x%5E2-2x%2B24%3C%3E+x%5E2-2x-24+
; so, here we will have complex solutions again