Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Write a polynomial $f(x)$ that satisfies the given conditions. Polynomial of lowest degree with zeros of $\frac{5}{3}$ (multiplicity $2$ ) and $-\frac{2}{3}$ (multiplicity $1$ ) and with $f(0)=-100$ .

View step-by-step help

View step-by-step help

Write a polynomial from its roots

Full solution

Q. Write a polynomial

f(x)

that satisfies the given conditions. Polynomial of lowest degree with zeros of

\frac{5}{3}

(multiplicity

2

) and

-\frac{2}{3}

(multiplicity

1

) and with

f(0)=-100

.

Identify Zeros and Factors: Identify the linear factors associated with the given zeros, considering their multiplicities. The zero $(\frac{5}{3})$ with multiplicity $2$ gives us the factor $(x - \frac{5}{3})^2$ , and the zero $-\left(\frac{2}{3}\right)$ with multiplicity $1$ gives us the factor $(x + \frac{2}{3})$ .
Write Factored Form: Write the polynomial in its factored form using the identified factors. The polynomial is $f(x) = (x - \frac{5}{3})^2 \cdot (x + \frac{2}{3})$ .
Expand and Simplify: Expand the polynomial to find the standard form. First, expand $(x - \frac{5}{3})^2$ to get $x^2 - (2\cdot\frac{5}{3})x + (\frac{5}{3})^2 = x^2 - (\frac{10}{3})x + \frac{25}{9}$ .
Distribute Terms: Next, expand the entire polynomial $f(x) = (x^2 - \frac{10}{3}x + \frac{25}{9}) * (x + \frac{2}{3})$ .
Combine Like Terms: Distribute the terms to get $f(x) = x^3 + \left(\frac{2}{3}\right)x^2 - \left(\frac{10}{3}\right)x^2 - \left(\frac{20}{9}\right)x + \left(\frac{25}{9}\right)x + \left(\frac{50}{27}\right)$ .
Adjust Constant Term: Combine like terms to simplify the polynomial. We get $f(x) = x^3 - \frac{8}{3}x^2 + \frac{5}{9}x + \frac{50}{27}$ .
Calculate New Constant: Now, we need to adjust the polynomial so that $f(0) = -100$ . Substitute $x = 0$ into the polynomial to find the current value of $f(0)$ . We get $f(0) = 0^3 - (8/3)\cdot0^2 + (5/9)\cdot0 + (50/27) = \frac{50}{27}$ .
Write Final Polynomial: To make $f(0) = -100$ , we need to adjust the constant term. Since $f(0)$ is currently $\frac{50}{27}$ , we need to subtract $\left(\frac{50}{27} + 100\right)$ from the constant term to get the desired value. Calculate the new constant term: $\left(\frac{50}{27}\right) - \left(\frac{50}{27} + 100\right) = -100$ .
Write Final Polynomial: To make $f(0) = -100$ , we need to adjust the constant term. Since $f(0)$ is currently $\frac{50}{27}$ , we need to subtract $\left(\frac{50}{27} + 100\right)$ from the constant term to get the desired value. Calculate the new constant term: $\left(\frac{50}{27}\right) - \left(\frac{50}{27} + 100\right) = -100$ .Write the final polynomial with the adjusted constant term. The polynomial is $f(x) = x^3 - \left(\frac{8}{3}\right)x^2 + \left(\frac{5}{9}\right)x - 100$ .

More problems from Write a polynomial from its roots

Question

Find the degree of this polynomial.

\newline

`5b^{10} + 3b^3`

\newline

Posted 2 months ago

Question

\newline

(9a + 5) - (2a + 4)

\newline

Posted 2 months ago

Question

Find the product. Simplify your answer.

\newline

-3v^2(v^2 - 9)

\newline

Posted 1 month ago

Question

Find the product. Simplify your answer.

\newline

(v - 3)(4v + 1)

\newline

Posted 2 months ago

Question

Find the square. Simplify your answer.

\newline

(3y + 2)^2

\newline

Posted 2 months ago

Question

Divide. If there is a remainder, include it as a simplified fraction.

\newline

(24t^2 + 36t) \div 6t

\newline

Posted 2 months ago

Question

Is the function

q(x) = x^6 - 9

even, odd, or neither?

\newline

\newline

\text{[[even][odd][neither]]}

Posted 2 months ago

Question

Find the product. Simplify your answer.

\newline

-3q^2(-3q^2 + q)

\newline

Posted 2 months ago

Question

Find the product. Simplify your answer.

\newline

(r + 3)(4r + 2)

\newline

Posted 2 months ago

Question

Find the roots of the factored polynomial.

\newline

(x + 7)(x + 4)

\newline

Write your answer as a list of values separated by commas.

\newline

x =

Posted 2 months ago