roots of polynomial equations help please

[lauren8754]

hey guys I would really appreciate it; if someone could show me the steps to figure out these problems.
Question: find the roots of each polynomial equation
1. x3 - 2x2 + 5x - 10 = 0 (the # 3 and 2 next to the X's are cubed and squared)


2. x3 = 5x2 + 7x - 35 =0


3. 4x3 + 16x2 - 22x - 10 =0

 

[galactus]

hey guys I would really appreciate it; if someone could show me the steps to figure out these problems.
Question: find the roots of each polynomial equation

1. \(\displaystyle x^{3}- 2x^{2} + 5x - 10 = 0\)

Why not use the ^ (SHIFT 6) to mean power?. That is what it's for. Anyway, Group: \(\displaystyle (x^{3}-2x^{2})+(5x-10)\)

Factor out common factors from each: \(\displaystyle x^{2}(x-2)+5(x-2)\)

See?. What is in the parentheses is the same. That is crucial for it to work.

\(\displaystyle (x^{2}+5)(x-2)\)...done.

Now, can you try the others?.

 

[Deleted member 4993]

hey guys I would really appreciate it; if someone could show me the steps to figure out these problems.
Question: find the roots of each polynomial equation
1. x3 - 2x2 + 5x - 10 = 0 (the # 3 and 2 next to the X's are cubed and squared)


2. x3 = 5x2 + 7x - 35 =0<<< This one is incorrect - please fix it


3. 4x3 + 16x2 - 22x - 10 =0

What have you tried?

What theorem/s about rational roots do you know?

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[soroban]

Hello, lauren8754!

The third one takes more work . . .

\(\displaystyle 3)\;\; 4x^3 + 16x^2 - 22x - 10 \:=\:0\)

\(\displaystyle \text{Divide by 2: }\;2x^3 + 8x^2 - 11x - 5 \:=\:0\)

The polynomial does not factor, so we use the Rational Roots Theorem.
. . And we find that \(\displaystyle x = -5\) is a root.

Using long division, we have: .\(\displaystyle (x+5)(2x^2-2x-1) \:=\:0\)

\(\displaystyle \text{Quadratic Formula: }\;x \:=\:\frac{2\pm\sqrt{12}}{2} \:=\:1 \pm \sqrt{3}\)


\(\displaystyle \text{The roots are: }\;x \;=\;-5,\;1\!+\!\sqrt{3},\;1\!-\!\sqrt{3}\)