# Can someone please help with this one? "write a polynomial equation of degree 4 with leading coefficient 1 that has roots at -2, -1, 3, and 4."?

##### 4 Answers
Apr 23, 2018

$\left(x + 2\right) \left(x + 1\right) \left(x - 3\right) \left(x - 4\right)$

#### Explanation:

This would give ${x}^{4}$ as your largest power, expand the brackets to write it as an equation.

Apr 23, 2018

${x}^{4} - 4 {x}^{3} - 7 {x}^{2} + 22 x + 24$

#### Explanation:

Expand $\left(x + 2\right) \left(x + 1\right) \left(x - 3\right) \left(x - 4\right)$

$\left(x + 2\right) \left(x + 1\right) = {x}^{2} + x + 2 x + 2 = {x}^{2} + 3 x + 2$

$\left({x}^{2} + 3 x + 2\right) \left(x - 3\right) = {x}^{3} - 3 {x}^{2} + 3 {x}^{2} - 9 x + 2 x - 6$

=${x}^{3} - 7 x - 6$

$\left({x}^{3} - 7 x - 6\right) \left(x - 4\right) = {x}^{4} - 4 {x}^{3} - 7 {x}^{2} + 28 x - 6 x + 24$

=${x}^{4} - 4 {x}^{3} - 7 {x}^{2} + 22 x + 24$

Apr 23, 2018

The polynomial is ${x}^{4} - 4 {x}^{3} - 7 {x}^{2} + 22 x + 24 = 0$

#### Explanation:

A polynomial equation of degree $4$ with leading coefficient $a$ and four roots $\alpha , \beta , \gamma , \delta$ is

$a \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) \left(x - \delta\right) = 0$

Hence, a polynomial equation of degree $4$ with leading coefficient $1$ and four roots $- 2 , - 1 , 3 , 4$ is

$1 \left(x - \left(- 2\right)\right) \left(x - \left(- 1\right)\right) \left(x - 3\right) \left(x - 4\right) = 0$

or $\left(x + 2\right) \left(x + 1\right) \left(x - 3\right) \left(x - 4\right) = 0$

or $\left({x}^{2} + 3 x + 2\right) \left({x}^{2} - 7 x + 12\right) = 0$

or ${x}^{4} - 7 {x}^{3} + 3 {x}^{3} + 12 {x}^{2} + 2 {x}^{2} - 21 {x}^{2} - 14 x + 36 x + 24 = 0$

or ${x}^{4} - 4 {x}^{3} - 7 {x}^{2} + 22 x + 24 = 0$

Apr 23, 2018

$f \left(x\right) = \left(x + 2\right) \left(x + 1\right) \left(x - 3\right) \left(x - 4\right)$

#### Explanation:

You can answer the question by looking at the requirements and fulfilling them step by step. A degree 4 polynomial will have ${x}^{4}$ in it somewhere, and 4 will be the largest number x is raised to in the equation. For example, $f \left(x\right) = {x}^{4} + {x}^{3} + {x}^{2} + x + 1$ is an equation with degree 4, since 4 is the largest number $x$ is raised to in the equation.

A leading coefficient of 1 means that the highest order term (${x}^{4}$ in this case) has the coefficient of 1. For example, $f \left(x\right) = 2 {x}^{4} + x$ has a leading coefficient of 2. You can think of $f \left(x\right) = {x}^{4}$ as having a "secret" 1 infront of the ${x}^{4}$ term. (i.e. $f \left(x\right) = 1 {x}^{4}$)

Lastly, having roots at -2, -1, 3, and 4 can be fulfilled by writing the polynomial in a way where it is easy to identify roots. A root of the equation is a value of $x$ where $f \left(x\right) = 0$. So for the equation $f \left(x\right) = \left(x - 5\right)$ the root is 5. Again, the roots for the equation $f \left(x\right) = \left(x - 2\right) \left(x - 3\right)$ are 2 and 3, since we have $f \left(2\right) = 0$ and $f \left(3\right) = 0$.

So all put together we can choose the equation $f \left(x\right) = \left(x + 2\right) \left(x + 1\right) \left(x - 3\right) \left(x - 4\right)$ which has roots at -2, -1, 3, and 4. We can expand the equation to $f \left(x\right) = {x}^{4} - 4 {x}^{3} - 7 {x}^{2} + 22 x + 24$ to double check that the equation has a leading coefficient of 1.