differential equation

Question

Find the value of a for which the polynomial x^5-ax²-ax+1has −1 as a root with multiplicity at least 2.

Answer 1

We have f(-1) = (-1)⁵ - a(-1)² - a(-1) + 1 = 0 or -1 - a + a +1 = 0 and 0 =0 true for any a;

and f'(-1) = 5(-1)⁴ -2a(-1) - a = 0, or 5 +2a - a = 0; a = -5.

Answer: a = -5 or polynomial x⁵ + 5x² + 5x + 1 has -1 as zero of multiplicity at least 2.

Answer 2

By using the product rule, D((x-c)^kg(x))=k(x-c)^k-1g(x)+(x-c)^kg'(x)=(x-c)^k-1(kg(x)+(x-c)g'(x)), we see that if c is a repeated root of f(x) of order k, then it is also a repeated root of f'(x) of order k-1. This gives us a quick way to test if -1 is a repeated root without factoring: we need f(-1)=f'(-1)=0.

So let f(x)=x⁵-ax²-ax+1, so that f'(x)=5x⁴-2ax-a. f(-1)=-1-a+a+1=0. f'(-1)=5+2a-a=5+a. The only way to make this 0 is if a=-5.

Of course, we can approach this without calculus too. By grouping x⁵-ax²-ax+1=(x⁵+1)-a(x²+x)=(x+1)(x⁴-x³+x²-x+1)-ax(x+1)=(x+1)(x⁴-x³+x²-x+1-ax), and for -1 to be a repeated root, it must be a zero of the second factor. Plugging in -1 to that factor, we obtain 1+1+1+1+1+a=5+a. Setting this equal to 0, we get a=-5.

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As an aside, one might ask "Other than -1, what other values can be double roots of f(x) for some choice of a?" By playing around with a graphing calculator (e.g., https://www.desmos.com/calculator/ijhxeoaipi), one can see that in addition to -1 being a double root when a=-5, there is a double root close to 0.9 when a is about 0.9 and another close to -0.5 when a is about -3.8. But how might one find them?

Suppose we had a value of a such that f(x) had a double root. Then f(x) and f'(x) would share a root, c. However, we can take the equations f(c)=0 and f'(c)=0 and solve them for a:

a=(c⁵+1)/(c²+c) and a=(5c⁴)/(2c+1)

Setting these two values of a equal to each other and simplifying, we get

(c+1)²(3c⁴-2c³+c²-1)=0

While we cannot solve this equation exactly, we can get numerical solutions, c≈0.837, -0.5677, and plugging into either of the above equations for a would give the approximate value of a need to get these as double roots.

Answer 3

By substituting -1 into the polynomial, we see that -1 is a root irrespective of the value of a. For -1 to be a root of multiplicity at least 2, -1 also has to be a root of the derivative, 5x^4 -2ax -a. Substituting -1 into the derivative, we have:

0 = 5 + a, i.e., a = -5.

Answer 4

You can verify that this polynomial has -1 as a root for all "a" by plugging x=-1 into the equation:

(-1)^5 - a (-1)^2 -a (-1) +1 = -1 -a +a +1 =0

So then you can factor out (x+1) from this. I find it simplest to consider the x^5+1 and the -a x^2 -a x separately as x=-1 is a root of both.

(x^5 + 1) = (x+1) (x^4 - x^3 + x^2 - x +1) and -ax^2 - ax = (x+1) (-ax).

So the polynomial reduces to (x+1) [x^4 - x^3 + x^2 - x - ax + 1].

The polynomial has x=-1 as a root with multiplicity at least 2 if and only if the expression in square brackets above has x=-1 as a root. So plugging x=-1 into this we have

(-1)^4 - (-1)^3 + (-1)^2 - (-1) - a (-1) + 1 = 1+1+1+1+a+1 = 5+a =0. Therefore a = -5.

Answer 5

For -1 to be a root, use Synthetic Division or regular polynomial division to get:

[ x⁵ - ax² - ax + 1 ] / (x + 1) = x⁴ - x³ + x² - (1+a)x + 1

Let f(x) = x⁴ - x³ + x² - (1+a)x + 1. Then f(-1) = 0 to make -1 a double root.

The remainder is zero only when a = 1. Thus we have

f(-1) = 1 - 1 + 1 - (1+a) + 1 = 1 - a = 0 only when a = 1.

Answer 6

Since -1 is a double root of the polynomial, it must also be a root of the first derivative of the polynomial. The first derivative is 5x⁴ - 2ax -a. So we must have 0 = 5(-1)⁴ - 2a(-1) - a = a + 5, and hence a = -5 is the only possible answer.

Answer 7

If -1 is a root, then (x + 1) is a factor. If -1 is a root with multiplicity of 2, then (x + 1)(x + 1) is a factor. Since (x + 1)(x + 1) = x² + 2x + 1, then x² + 2x + 1 is a factor and if we divide it into x⁵ - ax² - ax + 1 there will be a remainder of zero.

Using long division, we can divide x² + 2x + 1 into x⁵ - ax² - ax + 1 like this:

This is complicated but hopefully, you know how to do this. The bottom line is that the remainder is (5 + a)x + (5 + a). But we know that the remainder must be zero. So we can say 5 + a = 0 or a = -5