Unknown Polynomial Factorisation: Why does 3x^3+4x^2-x+m = (x-a)^2 (3x+b) ?

[Onigma]

Howdy Folks,

I was presented with the following polynomial equality question which I was able to complete fairly easily.

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[FONT=times\ new\ roman](a) \(\displaystyle \color{darkblue}\,x^3\, +\, x^2\, -\, 16x\, +\, k,\, k\, \in\, \mathbb{R},\,\) has two identical linear factors. Find k, and hence factorise the cubic into linear factors.[/FONT]

Below is my working I used to begin the problem. Note the first line of (x - a)^(2) * (x - b).

If \(\displaystyle \,P(x)\,\) has two identical linear factors, then, for some constants a and b, we have:

. . . . .\(\displaystyle \begin{align}x^3\, +\, x^2\, -\, 16x\, +\, k\, &=\, (x\, -\, a)^2\, (x\, -\, b)
\\\\
&=\, (x^2\, -\, 2ax\, +\, a^2)\, (x\, -\, b)
\\\\
&=\, x^3\, -\, bx^2\, -\, 2ax^2\, +\, 2abx\, +\, a^2x\, -\, a^2b
\\\\
&=\, x^3\, +\, (-2a\, -\, b)\, x^2\, +\, (2ab\, +\, a^2)\, x\, -\, a^2b\end{align}\)

I then continued this technique for a few more questions until I arrived at this final question where I was unable to get the correct answer.
Like prior, I began with (x - a)^(2) * (3x - b), taking note of the 3x as the coefficient of x^(3) was 3.

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[FONT=times\ new\ roman](c) \(\displaystyle \color{darkgreen}\,3x^3\, +\, 4x^2\, -\, x\, +\, m,\, m\, \in\, \mathbb{R},\,\) has two identical linear factors. Find the possible values of m, and find the zeroes of the polynomial in each case.[/FONT]

Looking at the worked solution for this problem though, I noticed that first line is not in fact (x - a)^(2) * (3x - b) but is rather (x - a)^(2) * (3x + b). Note how I was using a negative between the 3x and b, whereas the solution has a positive. What I don't understand is why there is a positive between for this question, even though the polynomial has all the same coefficient signs as the prior question (ie positives are the same and negatives are the same).

If \(\displaystyle \,P(x)\,\) has two identical linear factors, then, for some constants a and b, we have:

. . . . .\(\displaystyle \begin{align}3x^3\, +\, 4x^2\, -\, x\, +\, m\, &=\, (x\, -\, a)^2\, (3x\, +\, b)
\\ \\
&=\, (x^2\, -\, 2ax\, +\, a^2)\, (3x\, +\, b)
\\ \\
&=\, 3x^3\, +\, bx^2\, -\, 6ax^2\, -\, 2abx\, +\, 3a^2x\, +\, a^2b
\\ \\
&=\, 3x^3\, +\, (b\, -\, 6a)\, x^2\, +\, (3a^2\, -\, 2ab)\, x\, +\, a^2b\end{align}\)

After this lack of understanding of the first line, I am able to competently complete the rest of the question, it's just the very first line I don't understand.

Would someone please be able to explain to me firstly why it is correct to have a negative for the first question, but then why you need a positive in this second question. Also, how do you recognise that this final question requires a positive sign instead of negative.

Thanks Folks ; )

 

[Dr.Peterson]

Howdy Folks,

I was presented with the following polynomial equality question which I was able to complete fairly easily.

---

[FONT=times\ new\ roman](a) \(\displaystyle \color{darkblue}\,x^3\, +\, x^2\, -\, 16x\, +\, k,\, k\, \in\, \mathbb{R},\,\) has two identical linear factors. Find k, and hence factorise the cubic into linear factors.[/FONT]

Below is my working I used to begin the problem. Note the first line of (x - a)^(2) * (x - b).

If \(\displaystyle \,P(x)\,\) has two identical linear factors, then, for some constants a and b, we have:

. . . . .\(\displaystyle \begin{align}x^3\, +\, x^2\, -\, 16x\, +\, k\, &=\, (x\, -\, a)^2\, (x\, -\, b)
\\\\
&=\, (x^2\, -\, 2ax\, +\, a^2)\, (x\, -\, b)
\\\\
&=\, x^3\, -\, bx^2\, -\, 2ax^2\, +\, 2abx\, +\, a^2x\, -\, a^2b
\\\\
&=\, x^3\, +\, (-2a\, -\, b)\, x^2\, +\, (2ab\, +\, a^2)\, x\, -\, a^2b\end{align}\)

I then continued this technique for a few more questions until I arrived at this final question where I was unable to get the correct answer.
Like prior, I began with (x - a)^(2) * (3x - b), taking note of the 3x as the coefficient of x^(3) was 3.

---

[FONT=times\ new\ roman](c) \(\displaystyle \color{darkgreen}\,3x^3\, +\, 4x^2\, -\, x\, +\, m,\, m\, \in\, \mathbb{R},\,\) has two identical linear factors. Find the possible values of m, and find the zeroes of the polynomial in each case.[/FONT]

Looking at the worked solution for this problem though, I noticed that first line is not in fact (x - a)^(2) * (3x - b) but is rather (x - a)^(2) * (3x + b). Note how I was using a negative between the 3x and b, whereas the solution has a positive. What I don't understand is why there is a positive between for this question, even though the polynomial has all the same coefficient signs as the prior question (ie positives are the same and negatives are the same).

If \(\displaystyle \,P(x)\,\) has two identical linear factors, then, for some constants a and b, we have:

. . . . .\(\displaystyle \begin{align}3x^3\, +\, 4x^2\, -\, x\, +\, m\, &=\, (x\, -\, a)^2\, (3x\, +\, b)
\\ \\
&=\, (x^2\, -\, 2ax\, +\, a^2)\, (3x\, +\, b)
\\ \\
&=\, 3x^3\, +\, bx^2\, -\, 6ax^2\, -\, 2abx\, +\, 3a^2x\, +\, a^2b
\\ \\
&=\, 3x^3\, +\, (b\, -\, 6a)\, x^2\, +\, (3a^2\, -\, 2ab)\, x\, +\, a^2b\end{align}\)

After this lack of understanding of the first line, I am able to competently complete the rest of the question, it's just the very first line I don't understand.

Would someone please be able to explain to me firstly why it is correct to have a negative for the first question, but then why you need a positive in this second question. Also, how do you recognise that this final question requires a positive sign instead of negative.

Thanks Folks ; )

You don't need a positive sign; it isn't required, and either way will work. Both are correct. They just happen to have chosen a different way than you. (I myself might have chosen either 3(x - a)^(2) * (x - b) or 3(x + a)^(2) * (x + b). Using negative just makes it a tiny bit easier to get the roots at the end.)

What you need is to recognize when something doesn't matter. In this case, if you change the sign, it will just change the sign of your value of b, and nothing else. What you did is not wrong, and in fact is marginally more consistent.