Question

What is the completely factored form of f(x)=x^3+4x^2+7x+6?

f(x)=(x+2)(x−(−3+i10−−√))(x−(−3−i10−−√))
f(x)=(x+2)(x−(−1+i2√))(x−(−1−i2√)) f(x)=(x−2)(x−(−1+i2√))(x−(−1−i2√))
f(x)=(x−2)(x−(−3+i10−−√))(x−(−3−i10−−√))

Answer 1

we are given

`f(x)=x^3+4x^2+7x+6`

We will use rational root theorem to find factors

We can see that

Leading coefficient =1

constant term is 6

so, we will find all possible factors of 6

`6=\pm 1,\pm 2,\pm 3,\pm 6`

now, we will check each terms

At x=-2:

We can use synthetic division

we get

`f(-2)=0`

so, x+2 will be factor

and we can write our expression from synthetic division as

`f(x)=x^3+4x^2+7x+6=(x+2)(x^2+2x+3)`

`f(x)=(x+2)(x^2+2x+3)`

now, we can find factor of remaining terms

`x^2+2x+3=0`

we can use quadratic formula

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}`

`x=(-b\pm √(b^2-4ac))/(2a)`

we can compare our equation with quadratic equation

we get

`a=1,b=2,c=3`

now, we can plug these values

`x=(-2+√(2^2-4\cdot \:1\cdot \:3))/(2\cdot \:1)`

`x=-1+√(2)i`

`x=(-2-√(2^2-4\cdot \:1\cdot \:3))/(2\cdot \:1)`

`x=-1-√(2)i`

so, we get

`x=-1+√(2)i,\:x=-1-√(2)i`

so, we can write factor as

`x^2+2x+3=(x-(-1+√(2)i))(x-(-1-√(2)i))`

so, we get completely factored form as

`f(x)=(x+2)(x-(-1+√(2)i))(x-(-1-√(2)i))`...............Answer