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Q:Find the value of $a$ given that $x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$

No idea where to start, I was going to use the factor theorem but it didn't work out.

Question from year 10 Cambridge maths textbook

kjhg
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5 Answers5

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Probably too simplistic.

Write $$x^4-3x^3+3x^2+ax+2=(x^2+1)(x^2+Ax+B)$$ Expand and simplify; this would give $$(2-B)+x (a-A)+(2-B) x^2-(A+3) x^3=0$$ SInce this needs to be true for all $x$, set all coefficients equal to $0$. The first term leads to $B=2$, the last term to $A=-3$ and the second term to $a=A=-3$.

Claude Leibovici
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Divide your polynomial by $x^{2} + 1$ and set the remainder equal to zero.

Or, since you are only interested in the remainder, set $x^{2} = -1$ in the polynomial to get that the remainder is $$ (x^2)^{2}-3 x x^2+3 x^2+ax+2 = (-1)^{2} + 3 x - 3 + a x + 2 = (a + 3) x. $$ So $x^{2} + 1$ divides your polynomial iff $a = -3$.

Andreas Caranti
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$x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$ means $x=\pm i$ are the roots of the latter polynomial. Then you know what to do next.

user219967
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\begin{align} x^4-3x^3+3x^2+ax+2 &= (x^4 + x^2) - 3 \, (x^3 + x) + 2 \, (x^2 + 1) + (a+3) \, x \\ &= (x^2 + 1) \, \left( x^2 - 3 \, x + 2 + \frac{(a+3) \, x}{x^2 + 1} \right) \end{align} In order for this to be factored then it is required that $a = -3$.

Leucippus
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Don't make this question so hard. It is not necessary to factor x^2 + 1. Simply treat this as a long division problem, the type you did in grammar school, before calculators were invented. As a hint, it might be useful to think of the divisor as x^2 + 0x +1.

Another approach is to let x = 10. Then the problem reduces to 73A2 divided by 101, where A might be a negative integer. The problem then reduces to 72(10 + A)2 divided by 101 so that the second and fourth digits coincide. Eventually you will come to the equation 10 + A - 7 = 0.

Robert
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