Answer:
To find the roots of the polynomial P(x) = x^3 + 3x^2 + 3x + 2, we can follow the steps below:
List the 4 possible rational roots:
The possible rational roots of the polynomial P(x) are the divisors of the constant term 2, which are -1, 1, -2, and 2.
Plug each possible rational root into P(x) to determine which is a root of the polynomial:
Plugging in -1 for x, we get P(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 2 = -1 + 3 + 3 - 2 = 1.
Plugging in 1 for x, we get P(1) = 1^3 + 3(1)^2 + 3(1) + 2 = 1 + 3 + 3 + 2 = 9.
Plugging in -2 for x, we get P(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 2 = -8 + 12 + 6 - 2 = -2.
Plugging in 2 for x, we get P(2) = 2^3 + 3(2)^2 + 3(2) + 2 = 8 + 12 + 6 + 2 = 28.
Thus, only x = -2 is a root of the polynomial P(x).
Divide P(x) by the root you found in part 2 using synthetic division:
To divide P(x) by the root x = -2 using synthetic division.
Thus, the quotient of the division is x^2 + 5x + 3.Write the fully factored form of P(x):
The fully factored form of P(x) is (x + 2)(x^2 + 5x + 3).Use the quadratic formula to find the solutions of the quadratic factor:
To find the solutions of the quadratic factor x^2 + 5x + 3, we can use the quadratic formula:x = (-b +/- sqrt(b^2 - 4ac)) / (2a)where a = 1, b = 5, and c = 3. Plugging these values into the formula, we get:x = (-5 +/- sqrt(5^2 - 4(1)(3))) / (2(1))
= (-5 +/- sqrt(25 - 12)) / 2
= (-5 +/- sqrt(13)) / 2This gives us the two solutions x = (-5 + sqrt(13)) / 2 and x = (-5 - sqrt(13)) / 2.List the 3 solutions of the P(x):
The 3 solutions of the polynomial P(x) are x = -2, x = (-5 + sqrt(13)) / 2, and x = (-5 - sqrt(13)) / 2.
Step-by-step explanation:
Answer:
[tex]p(x)=(x+2)(x^2+x+1)[/tex]
[tex]x=-2[/tex]
[tex]x=\dfrac{-1 + \sqrt{3}\:i}{2}[/tex]
[tex]x=\dfrac{-1- \sqrt{3}\:i}{2}[/tex]
Step-by-step explanation:
Given polynomial:
[tex]p(x)=x^3+3x^2+3x+2[/tex]
Rational Root Theorem
If P(x) is a polynomial with integer coefficients and if p/q is a root of P(x), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x).
Possible p-values
Factors of the constant term: ±1, ±2
Possible q-values
Factors of the leading coefficient: ±1
Therefore, all the possible values of p/q:
[tex]\sf \dfrac{p}{q}=\dfrac{\pm 1}{\pm 1}, \dfrac{\pm 2}{\pm 1}=\pm1,\pm2[/tex]
Substitute each possible rational root into the function:
[tex]x=-1 \implies p(-1) =(-1)^3+3(-1)^2+3(-1)+2=1[/tex]
[tex]x=1 \implies p(1) =(1)^3+3(1)^2+3(1)+2=9[/tex]
[tex]x=-2 \implies p(-2) =(-2)^3+3(-2)^2+3(-2)+2=0[/tex]
[tex]x=2 \implies p(2) =(2)^3+3(2)^2+3(2)+2=28[/tex]
Therefore, x = -2 is a root of the polynomial since f(-2) = 0.
Divide the polynomial by the root using synthetic division:
[tex]\begin{array}{c|crrr}-2 & 1 & 3 & 3 & 2\\\cline{1-1} & \downarrow &-2&-2&-2\\ \cline{2-5} & 1&1&1&0\end{array}[/tex]
The bottom row (except the last number) gives the coefficients of the quotient. Therefore, the quotient is:
[tex]x^2+x+1[/tex]
So the fully factored form of p(x) is:
[tex]p(x)= (x+2)(x^2+x+1)[/tex]
[tex]\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}[/tex]
Therefore, the solutions to the quadratic factor are:
[tex]\implies x=\dfrac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}[/tex]
[tex]\implies x=\dfrac{-1 \pm \sqrt{-3}}{2}[/tex]
[tex]\implies x=\dfrac{-1 \pm \sqrt{3\cdot -1}}{2}[/tex]
[tex]\implies x=\dfrac{-1 \pm \sqrt{3} \sqrt{-1}}{2}[/tex]
[tex]\implies x=\dfrac{-1 \pm \sqrt{3}\:i}{2}[/tex]
The 3 solutions of p(x) are: