# Does anybody know how he got x^2+4x+5?

### 11 Answers

- L. E. GantLv 712 hours ago
(x - (-2 +i))(x-(-2 -i))

= ((x+2) - i)((x+2) + i)

= (x+2)^2 - i^2

= x^2 + 4x + 4 + 1

= x^2 + 4x + 5

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- PinkgreenLv 71 day ago
If an equation has a complex root like x=-2+i, it must

have its conjugate x=-2-i as a root. So,

(x+2-i)(x+2+i)=(x+2)^2-i^2=x^2+4x+5 is a factor of the given equation f(x)=0. Thus,

f(x)=(x+2-i)(x+2+i)(x^2+6x+9)=0

=>

(x+2-i)(x+2+i)(x+3)^2=0

=>

x=-2+i

x=-2-i

x=-3 (repeated roots)

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- PhilipLv 62 days ago
Since x = -2-i and x = -2+i are both zeros of 4th degree

polynomial function f(x) = x^4 +10x^3 +38x^2 +66x +45,

[x-(-2-i)][x-(-2+i)]=(x+2+i)(x+2-i)=(x+2)^2 -i^2=x^2+4x+5 is

a factor of f(x). That's how he got it! Dividing f(x) by

x^2+4x+5 gave the quadratic g(x) = x^2+6x+9 = (x+3)^2.

The 2 zeros of g(x) are x = -3{multiplicity 2} which are the

final 2 zeros of f(x).

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- Fazaldin ALv 72 days ago
find roots for : x^4 +10x^3 +38x^2 +66x +45 = 0, ------------------- [1]

Roots are: x1 = -2+i, x2 = -2-i, x3 =?, and x4 = ?

x1 * x2 = (x+2-i)(x+2+i) = x^2 +x(2-i +2+i) + (2-i)(2+i) = x^2+4x +5

So,

x^4 +10x^3 +38x^2 +66x +45 = (x^2+4x+5)(x^2 +bx +c) = 0,

Or,

x^2(x^2+bx+c) +4x(x^2+bx+c) +5(x^2+bx+c) = 0,

Or,

x^4 +x^3(b+4) +x^2(c+4b+5) +x(4c+5b) +5c = 0, ..................[2]

Comparing [1] & [2],

b+4 = 10, 4b+c+5 = 38, 4c+5b = 66, and 5c = 45

b = 6, 24+c+5 = 38, 4c+5b = 66, and c + 9,

Or,

b=6, c= 9,

Hence,

x^4 +10x^3 +38x^2 +66x +45 = (x^2+4x+5)(x^2 +6x +9) = 0,

For,

x^2+4x+5 = 0, x= -2+i, & x = -2-i,

and for,

x^2 +6x +9 = 0, x = -3, and x = -3

The roots are:

x= -2+i, x = -2-i, x = -3, and x = -3. ------------------ Anwr

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- sepiaLv 73 days ago
x^4 + 10x^3 + 38x^2 + 66x + 45 = 0

(x + 3)^2 (x^2 + 4 x + 5) = 0

Real solution:

x = -3

Complex solutions:

x = -2 - i

x = -2 + i

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- PuzzlingLv 73 days ago
We are told one solution and we can infer the conjugate:

x = -2 + i

x = -2 - i

Rewrite those as expressions equal to 0:

x + 2 - i = 0

x + 2 + i = 0

Now group x+2:

(x + 2) - i = 0

(x + 2) + i = 0

Finally multiply them together:

[(x + 2) - i][(x + 2) + i]

It's a difference of squares:

(x + 2)² - i²

= x² + 4x + 4 - (-1)

= x² + 4x + 5

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- PuzzlingLv 72 days agoReport
I was explaining why the person wrote down the conjugate pair. It was not given originally.

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- 3 days ago
yes,

root of x = -2 + i

and x = -2 -i

means that

(x+ 2 -i)(x+ 2 +i) are factors

use distributive rule to

=x^2 + 2x +ix + 2x + 4 + 2i -ix -2i - (i^2) = 0

= x^2 + (2x + 2x -ix + ix ) + 4 - (-1) = 0

= x^2 + 4x + 4 + 1

=x^2 + 4x + 5 (that's how

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- KrishnamurthyLv 73 days ago
x^4 + 10x^3 + 38x^2 + 66x + 45 = 0

(x + 3)^2 (x^2 + 4 x + 5) = 0

Real solution:

x = -3

Complex solutions:

x = -2 - i

x = -2 + i

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- ted sLv 73 days ago
you are told that - 2 ± i are roots ===> (x - (-2 ))² + 1 is a factor = x² + 4x + 5

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- PaladinLv 73 days ago
change the two solutions into factors then multiply the two factors together

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