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."?

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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."?

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Answer 1 · 2018-04-23T15:38:51

$(x + 2) (x + 1) (x - 3) (x - 4)$ $(x+2)(x+1)(x-3)(x-4)$

Explanation:

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

Answer 2 · 2018-04-23T16:22:22

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

Explanation:

Expand $(x + 2) (x + 1) (x - 3) (x - 4)$ $(x+2)(x+1)(x-3)(x-4)$

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

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

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

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

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

Answer 3 · 2018-04-23T16:37:31

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

Explanation:

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

$a (x - α) (x - β) (x - γ) (x - δ) = 0$ $a(x-alpha)(x-beta)(x-gamma)(x-delta)=0$

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

$1 (x - (- 2)) (x - (- 1)) (x - 3) (x - 4) = 0$ $1(x-(-2))(x-(-1))(x-3)(x-4)=0$

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

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

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

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

Answer 4 · 2018-04-23T21:08:41

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

Explanation:

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

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

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 $x$ $x$ where $f (x) = 0$ $f(x) = 0$ . So for the equation $f (x) = (x - 5)$ $f(x)=(x-5)$ the root is 5. Again, the roots for the equation $f (x) = (x - 2) (x - 3)$ $f(x)=(x-2)(x-3)$ are 2 and 3, since we have $f (2) = 0$ $f(2) = 0$ and $f (3) = 0$ $f(3) = 0$ .

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

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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

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