# Determine a third-degree polynomial with integer coefficients and roots 1/2, 1/3, 1/4

#### Aion

##### New member
Determine a third-degree polynomial with integer coefficients and roots 1/2, 1/3 and 1/4.

Well, I know that we can write the polynomial as

$$\displaystyle (x-1/2)(x-1/3)(x-1/4)=0$$

$$\displaystyle x^3 - \frac{13 x^2}{12} + \frac{3 x}{8} - \frac{1}{24} = 0$$

$$\displaystyle \frac{1}{24}(24 x^3 - 26 x^2 + 9 x - 1) = 0$$

$$\displaystyle 24 x^3 - 26 x^2 + 9 x - 1 = 0$$

Is this the form they are after?

Also, does anyone have any good math books recommendations?

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#### ksdhart2

##### Full Member
Well, if you're doubting your solution, you can always check it to see if it meets all of the criteria.

Determine a third-degree polynomial...
Let's quickly remind ourselves of the definition of "degree" in this context. The degree of a polynomial is the highest power involved in the expression. A third degree polynomial would need the highest power to be x^3, which we have. $$\displaystyle \checkmark$$

...with integer coefficients...
The coefficients are the numbers out front of the x terms. 24, 26, 9, and 1 are integers, so we're good here too. $$\displaystyle \checkmark$$

...and roots 1/2, 1/3 and 1/4
The factor theorem tells us that if x = q is a root of a polynomial, then (x - q) must be a factor of that polynomial, and vice versa. $$\displaystyle 24x^3 - 26x^2 + 9x - 1 = 24\left(x - \frac{1}{2}\right)\left(x - \frac{1}{3}\right)\left(x - \frac{1}{4}\right)$$, so we have the right roots. $$\displaystyle \checkmark$$

The only small quibble I have is that what you wrote is an equation, whereas you're asked for a polynomial. Strictly speaking, the polynomial is only the part to the left of the equals sign.

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#### Dr.Peterson

##### Elite Member
Determine a third-degree polynomial with integer coefficients and roots 1/2, 1/3 and 1/4.

Well, I know that we can write the polynomial as

$$\displaystyle (x-1/2)(x-1/3)(x-1/4)=0$$

$$\displaystyle x^3 - \frac{13 x^2}{12} + \frac{3 x}{8} - \frac{1}{24} = 0$$

$$\displaystyle \frac{1}{24}(24 x^3 - 26 x^2 + 9 x - 1) = 0$$

$$\displaystyle 24 x^3 - 26 x^2 + 9 x - 1 = 0$$

Is this the form they are after?
Yes.

You can make the work a little easier by starting with integer coefficients for the factors, $$\displaystyle (2x-1)(3x-1)(4x-1) = (6x^2 - 5x + 1)(4x - 1) = 24x^3 - 26x^2 + 9x - 1$$.

#### HallsofIvy

##### Elite Member
Yes, that is correct.

#### Aion

##### New member
Yes.

You can make the work a little easier by starting with integer coefficients for the factors, $$\displaystyle (2x-1)(3x-1)(4x-1) = (6x^2 - 5x + 1)(4x - 1) = 24x^3 - 26x^2 + 9x - 1$$.
Hey, thanks. This actually makes it a lot easier!

#### Jomo

##### Elite Member
Determine a third-degree polynomial with integer coefficients and roots 1/2, 1/3 and 1/4.

Well, I know that we can write the polynomial as

$$\displaystyle (x-1/2)(x-1/3)(x-1/4)=0$$

$$\displaystyle x^3 - \frac{13 x^2}{12} + \frac{3 x}{8} - \frac{1}{24} = 0$$

$$\displaystyle \frac{1}{24}(24 x^3 - 26 x^2 + 9 x - 1) = 0$$

$$\displaystyle 24 x^3 - 26 x^2 + 9 x - 1 = 0$$

Is this the form they are after?

Also, does anyone have any good math books recommendations?
To go 1 step further from what Dr P said, notice that after you multiply the 1st factor by 2 and the 2nd factor by 3 and the 3rd factor by 4, you multiplied the whole expression by 2*3*4 which is 24. The same 24 that you eventually used. As you said, it is much easier to multiply 1st. The number 0 is cool because no matter what you multiply it by it stays 0. One day when you study advanced algebra you will see the beauty in the number 0.