$\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}= \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1$

7 0

3 years ago

WILL GIVE BRAINLIEST PLEASE ANSWER ASAP

Wittaler [7]

Answer:

Circle man ez

Step-by-step explanation:

5 0

3 years ago

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1. On the first day of ticket sales, the school sold a total of 25 adult and child tickets. Each adult

faltersainse [42]

13 adults and 10 children

7 0

3 years ago

What is the fifth term of the geometric sequence a1=120, a2=36, a3=10.8, a6=0.2916?

iragen [17]

First we need to find the common ratio by dividing the second term by the first term. 36/120 = 3/10

an = a1 * r^(n-1)
n = term to find = 5
a1 = first term = 120
r = common ratio = 3/10
now we sub and solve
a5 = 120 * 3/10^(5 - 1)
a5 = 120 * 3/10^4
a5 = 120 * .0081
a5 = 0.972 <=== fifth term

or we could have just done this......since we multiply by 3/10 to find the next number...
a3 = 10.8......10.8 * 3/10 = 3.24 <== 4th term
3.24 * 3/10 = 0.972 <== fifth term

5 0

3 years ago

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