Integer Coefficients

By the Rational Root Theorem, we must have $2|e$ , $3|e$ , and $5|e$ . This means $e \ge 30$ , achievable with

$p(x) = (x+2)(x-3)(3x-5)(2x+1)$

For the polynomial of degree 4, we are given the four roots.

$(x+2)(x-3)(2x+1)(3x-5) = ax^{4}+bx^{3}+cx^2+dx+e$

Since we only need the e term, we can multiply the second term in each two-termed part of the expression.

$(2)(-3)(1)(-5) = 30$

Let ax^4+bx^3+cx^2+dx+e = k(x+2)(x-3)(x-5÷3)(x+1/2) = k(x^4-13x^3÷3+47x÷6-17x^2÷3+5) For e to be minimum and the coefficients to be integer k =6. So e = 6×5 = 30.

Given the roots, we can conclude the following values of $r$ (or $x$ , since they're interchangeable):

$\displaystyle x_1a=-2;x_2=3;x_3=\frac{5}{3};x_4=-\frac{1}{2}$

By multiplying the denominators and transposing the constants, we find that

$\displaystyle ax^4+bx^3+cx^2+dx+e=(x+2)(x-3)(3x-5)(2x+1)$

Since $e$ can only be found by multiplying the constants:

$e=(2)(-3)(-5)(1)=\boxed{30}$

the roots are given namely;

x = (-2), (3), ( $\frac{5}{3}$ ), and (- $\frac{1}{2}$ )

it can be written as

(x + 2)(x - 3)(3x - 5)(2x + 1) = 0

finding the "e", we can just multiply the terms without a variable:

(2)(-3)(-5)(1) = 30

Since the polynomial:

$P(x)= ax^4+bx^3+cx^2+dx+e$

is required to have integer coefficients and the roots are indeed rational numbers, we may apply the rational roots theorem , which states that the any rational root $\frac{\alpha}{\beta}$ is such that $\alpha$ divides $e$ and $\beta$ divides $a$ . From here, we go through the roots and determine that $2$ , $3$ , and $5$ must divide $e$ . The lowest possible common factor is then, $30$ , which is the solution.

Since $ax^{4}$ + $bx^{3}$ + $cx^{2}$ + $dx$ + $e$ has roots at $-2, 3, \frac {5}{3},$ and $-\frac {1}{2}$

We could say that ( $x+2$ )( $x-3$ )( $3x-5$ )( $2x+1$ ) $=0$

Then,

( $x^{2}-x-6$ )( $6x^{2}-7x-5$ ) $=0$

$6x^{4}-55x^{3}-34x^{2}+47x+30$ $=0$

Thus, $e=30$

• We know that the roots of this can be expressed as:

• * b/a = r1+r2+r3..r5 *

In order for this to have integer coefficients, *a must be a multiple of both 3 and 2, in order to cancel out 5/3 and -1/2 .*

• We also know that e can be expressed as the multiple of all the roots of the polynomial.

• Henceforth, it is easy to conclude that e/a = 5.

• Since a is a multiple of 6 (in its lowest form), the answer is * *e=6 5 = 30.**

In order to solve this problem, you have to realize that the roots of the polynomial were found by setting each factor equal to 0. So if a root is -2, then it came from the factor (x + 2). By using this, the factors of the polynomial can be found to be (x + 2)(x - 3)(x - 5/3)(x + 1/2). By multiplying this out, you find that e equals 30.

Let the equation be f(x), it can be rewritten as such

(x+2)(x-3)(x-5/3)(x+1/2) = 0

This would give u all the roots in the question, however the polynomial has to have integer coefficients, expanding with a fraction inside would result in some coefficients to be a fraction.

Thus written as,

(x+2)(x-3)(3x-5)(2x-1)=0

As we are looking for the constant, there is no need to expand the whole equation, just take the constants in the equation and multiply them together.

So we have,

2 x -3 x -5 x 1= 30

Hope this helps!

Since -2 , 3 , 5/3 , and -1/2 are the roots of the equation, this implies that $({x}+{2})({x}+{3})({3x}-{5})({2x}+{1})={0}$ Expanding these brackets we get $({x^2}+{6x^2}+{6})({6x^2}-{7x}-{5})={0}$ $({6x^4}+{29x^3}-{11x^2}-{72x}-{30})={0}$ But ${ax^4}+{bx^3}+{cx^2}+{dx}+{e}= {6x^4}+{29x^3}-{11x^2}-{72x}-{30}$

therefore the smallest positive integer of e equals
$\boxed{30}$

Let p,q,r,s be the roots of the equation $$ax^4 +bx^3 +cx^2 +dx +e$$ Where $$p=-2, q=3, r=\frac {5}{3}, s=\frac {-1}{2} $$ As we know, product of roots of equation= $$\frac {e}{a}$$ Hence, $$p * q * r * s=\frac {e}{a} $$ $$\Rightarrow \frac {e}{a}= \frac {30}{6} $$ Again, sum of roots=$$\frac {-b}{a}$$ $$\Rightarrow \frac {-b}{a}= p+q+r+s $$ $$\Rightarrow \frac {-b}{a}= \frac {13}{6}$$ Again, sum of product of two roots taken at a time=$$\frac {c}{a}$$ $$\Rightarrow p * q+q * r+r * s+s * p= \frac {c}{a} $$ $$\Rightarrow \frac {c}{a}= \frac {-5}{6}$$ Again, sum of product of three roots taken at a time= $$\frac {-d}{a}$$ $$\Rightarrow \frac {-d}{a}=p * q * r + q * r * s + r * s * p + s * p * q $$ $$\Rightarrow \frac {-d}{a}=-7$$ $$\Rightarrow \frac {-d}{a}=\frac {42}{6} $$ Hence we get; $$ a=6,b=-13,c=-5,d=42, \boxed {e=30}$$

With Vietha's theorem that e/a = -2 . 3 . 5/3 . (-1/2)= 5, Because it consist 2,3, and 5 so, 5 x 3 x 2 = 30.

(x + 2)(x - 3)(3x - 5)(2x + 1) .... when we expand this, the constant term 'e' is the product of all constant terms in the brackets.