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I was trying to factor this polynomial:

$x^3 + x^2 - 16x + 20$

There are four options in this question:

  1. (a) It could be factored in the following form: $(x^2 + b)(x+c)$;
  2. (b) It could be factored in the following form: $(x+b)(x+c)(x+d)$, assuming that $b \neq c \neq d$
  3. (c) It could not be factored.
  4. (d) It could be factored in the following form: $(x+b)^2 (x+c) $

Here's how I've tried to do it: I've tried to factor by grouping the x, therefore I've obtained: $x(x^2 + x - 16) + 20$. Now, I've put the $x$ and the $20$ together: $(x+20)(x^2 + x - 16)$. Then, I've tried to factor the second term: $(x+20)(x-16)(x+1)$. So, the answer would be "b", according to this algorithm.

I've completed the test (it's a simulation for the admission test I'm going to do), I submit the answers, and I've noticed that this question isn't correct.

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    $\begingroup$ Doesn't it factor as $(x-2)^2(x+5)$? $\endgroup$
    – saulspatz
    Commented Nov 25, 2020 at 7:19
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    $\begingroup$ You cannot factor out the $(x+20)$ as you have done. There is no common factor of $(x+20)$ between $x^2 + x -16$ and $20$. Hint on what to do: Guess at a root of the polynomial and perform synthetic division. $\endgroup$ Commented Nov 25, 2020 at 7:19
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    $\begingroup$ Theorem: every polynomial to be factorized in a homework question has small, integer roots :) $\endgroup$
    – user438666
    Commented Nov 25, 2020 at 15:42

3 Answers 3

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As @Fernis pointed out in the comments,

You cannot factor out the $(x+20)$ as you have done. There is no common factor of $(x+20)$ between $x^2+x−16$ and $20$.

Using the Rational Root Theorem, you can know that the possible rational roots are $\pm 1, \pm2, \pm4, \pm5, \pm10, \pm20$.

Through inspection and polynomial/synthetic division, you can get $(x-2)^2(x-5)$, as @saulspatz said. Therefore, (d) is your answer.

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The easiest to try is (d), because it says that the polynomial has a double root. We will look for a root of the derivative and check if it cancels the polynomial.

$$3x^2+2x-16=0\iff x=2\text{ or }x=-\dfrac83.$$

Now $p(2)=0$, bingo !

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    $\begingroup$ Good solution, but not sure if OP understands calculus $\endgroup$
    – KingLogic
    Commented Nov 25, 2020 at 20:22
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This one is easy to factor. Let's see how.

Start by plugging in $x=0,1,-1,2$ and so on.

You will find by inspection that $x=2$ is a zero of the polynomial. Therefore,$(x-2)$ is its factor.

Now factor the polynomial in such a way that $(x-2)$ gets common.

$$x^3+x^2-16x+20$$ $$=x^3-2x^2+3x^2-6x-10x+20$$ $$=x^2(x-2)+3x(x-2)-10(x-2)$$ $$=(x-2)(x^2+3x-10)$$ $$=(x-2)(x^2+5x-2x-10)$$ $$=(x-2)[x(x+5)-2(x+5)]$$ $$=(x-2)^2(x+5)$$

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