Find the value of a^4-a^3+a^2+2 when a^2+2=2a

[sudarshan]

Find the value of a⁴-a³+a²+2 when a²+2=2a
ans:0
Factorize
(x-1)(x-2)(x+3)(x+4)+7

[perfectmath]

Find the value of a^4-a^3+a^2+2 when a^2+2=2a
ans:0
Factorize
(x-1)(x-2)(x+3)(x+4)+7

[tex](x-1)(x-2)(x+3)(x+4)+7=x^4+4x^3-7x^2-22x+31[/tex].

[MM]

[tex]a^4-a^3+a^2+2=a^4-a^3+2a=a(a^3-a^2+2)=a(a^3+2-2a+2)=a(a(a^2-2)+4)=a(a(2a-4)+4)[/tex] [tex]a(a(2a-4)+4)=a(2a^2-4a+4)=2a(a^2+2-2a)=0[/tex] because [tex]a^2+2-2a=0[/tex]
Now try to find [tex]a^4-a^3+a^2+2[/tex] if [tex]a^2+a+1=0[/tex].

[sudarshan]

Thank you MM
If the answer is 0, i have solution.

[MM]

The answer is 0, indeed. Post your solution.

[Guest]

Check this out
a^4-a^3+a^2+2
=a^2(a^2-a+1)+2
=a(-a-a)+2
=a(-2a)+2
=2-2a^3
=2(1-a^3)
=2(1-a)(1+a+a^2)
=2(1-a)(0)
=0

[sudarshan]

Check this out
a^4-a^3+a^2+2
=a^2(a^2-a+1)+2
=a(-a-a)+2
=a(-2a)+2
=2-2a^3
=2(1-a^3)
=2(1-a)(1+a+a^2)
=2(1-a)(0)
=0

It's me. please believe me, i was so exited to post my solution that i forget to log in.

[Jessica]

Yeah I also got the solution!!!!!!!

[Guest]

a^4 - a^3 + a^2 + 2 = a^2 (a^2 - a + 1) +2
= a^2 (2a - 2 - a + 1) + 2
= a^2 (a - 1) + 2
= (2a - 2)(a - 1) + 2
= 2 a^2 - 4 a + 4
= 2(a^2 - 2 a + 2)
but a^2 + 2 = 2 a


a^4 - a^3 + a^2 + 2 = 2 (2a - 2a)

= 0

[Mathmaven53]

a^4 - a^3 + a^2 + 2 = a^2 (a^2 - a + 1) +2
= a^2 (2a - 2 - a + 1) + 2
= a^2 (a - 1) + 2
= (2a - 2)(a - 1) + 2
= 2 a^2 - 4 a + 4
= 2(a^2 - 2 a + 2)
but a^2 + 2 = 2 a


a^4 - a^3 + a^2 + 2 = 2 (2a - 2a)

= 0

[HallsofIvy]

Working "the other way around", if a^2+ 2= 2a then a^2- 2a+ 2= 0. Multiplying by a^2, a^4- 2a^3+ 2a^2= 0.
Then (a^4- 2a^3+ 2a^2)+ (a^3- a^2+ 2)= a^4- a^3+ a^2+ 2= 0+ 0= 0.

[Guest]

Since others have given every clever method, here is an unclever, brute strength, method:
[tex]A^2+ 2= 2A[/tex]
[tex]A^2- 2A= -2[/tex]
[tex]A^2- 2A+ 1= (A- 1)^2= -1[/tex]
[tex]A- 1= \pm i[/tex]
[tex]A= 1\pm i[/tex].

If [tex]A= 1+ i[/tex], [tex]A^2= 2i[/tex], [tex]A^3= -2+ 2i[/tex], and [tex]A^4= -4[/tex].

So [tex]A^4-A^3+A^2+2= -4+ 2- 2i+ 2i+ 2= 4[/tex].

[ammornil]

[tex]a^{2}-2a+2=0 \Rightarrow (a-2)^{2}=0 \Rightarrow a=2[/tex]

[tex]a^{4}-a^{3}+a^{2}+2=2^{4}-2^{3}+2^{2}+2=16-8+4+2=14[/tex]

if [tex]a=0[/tex] then [tex]a^{2}+2 \ne a[/tex]

[GeradHum]

First, from a2+2=2aa^2 + 2 = 2aa2+2=2a, we get:
a2−2a+2=0a^2 - 2a + 2 = 0a2−2a+2=0
Now plug into the expression:
a4−a3+a2+2=0(after simplifying using the equation above)a^4 - a^3 + a^2 + 2 = 0 \quad \text{(after simplifying using the equation above)}a4−a3+a2+2=0(after simplifying using the equation above)
Answer: 0
For the second part:
(x−1)(x−2)(x+3)(x+4)+7=(x2+2x−3)(x2+7x+12)+7(x - 1)(x - 2)(x + 3)(x + 4) + 7 = (x^2 + 2x - 3)(x^2 + 7x + 12) + 7(x−1)(x−2)(x+3)(x+4)+7=(x2+2x−3)(x2+7x+12)+7
Multiply and simplify — the expression is factorable, but it requires expansion.