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

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

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

we can compare our equation with quadratic equation

we get

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

now, we can plug these values

$x=\frac{-2+\sqrt{2^2-4\cdot \:1\cdot \:3}}{2\cdot \:1}$

$x=-1+\sqrt{2}i$

$x=\frac{-2-\sqrt{2^2-4\cdot \:1\cdot \:3}}{2\cdot \:1}$

$x=-1-\sqrt{2}i$

so, we get

$x=-1+\sqrt{2}i,\:x=-1-\sqrt{2}i$

so, we can write factor as

$x^2+2x+3=(x-(-1+\sqrt{2}i))(x-(-1-\sqrt{2}i))$

so, we get completely factored form as

$f(x)=(x+2)(x-(-1+\sqrt{2}i))(x-(-1-\sqrt{2}i))$...............Answer