# Math IB HL math portfolio type I - polynomials

Diana Herwono D 0861 006

1. Let P(x) = a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0.

Using the sums of (a5 + a3 + a1) and (a4 + a2 + a0), determine whether

P(1) = 0?  and P(-1) = 0?

Examine these examples:

(1) P(x) = x5 - 3x4 + 2x3 + 4x2 + 6x - 10

(2) P(x) = x5 - 3x4 + 2x3 - 4x2 + 6x + 10

(3) P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x + 10

(4) P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x - 10

## What is your conclusion for the general case when

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x1 + a0?

Solution:

(1)         P(x) = x5 - 3x4 + 2x3 + 4x2 + 6x - 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = -9

P(1) = 0

P(-1) = -18

(2)         P(x) = x5 - 3x4 + 2x3 - 4x2 + 6x + 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = 3

P(1) = 12

P(-1) = -6

(3)        P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x + 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = 9

P(1) = 18

P(-1) = 0

(4)        P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x - 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = -11

P(1) = -2

P(-1) = -20

Conclusion:

From the above examples, we see that when:

a5 + a3 + a1 = -(a4 + a2 + a0)                 P(1)= 0

a5 + a3 + a1 = a4 + a2 + a0                 P(-1)= 0

Therefore for the general case,

if an-1 + an-3 + … + a1 = -(an + … + a2 + a0)         then P(1) = 0

if an-1 + an-3 + … + a1 = an + … + a2 + a0         then P(-1) = 0

1. There is a conclusion that states:

If an integer k is a zero of a polynomial with integral coefficients, then k must be a factor of the constant term of the polynomial.

To understand this conclusion, study the function P(x) = a3x3 + a2x2 + a1x + a0

and suppose that P(k) = 0. Can you see that k must be a factor of a0?

Solution:

P(x) = a3x3 + a2x2 + a1x + a0

P(k) = a3k3 + a2k2 + a1k + a0

Suppose P(k) = 0 then

a3k3 + a2k2 + a1k + a0 = 0

a0 = -(a3k3 + a2k2 + a1k)

= k (-a3k2 - a2k - a1)

Therefore, k must be a factor of a0.

1. There is a conclusion that states:

If a0 = 0, then P(x) = anxn + an-1xn-1 + … + a1x + a0 has integral coefficients, and the rational number k/m in lowest terms is a zero of P(x), then k must be a factor of a0, and m must be a factor of an.

## To understand this conclusion, study the function P(x)= a3x3+a2x2+a1x+a0, and suppose that P(k/m) = 0.  Can you see that k must be a factor of a0, and m must be a factor of a3?

Solution:

P(x) = a3x3 + a2x2 + a1x + a0

P(k/m) = a3(k/m)3 + a2(k/m)2 + a1(k/m) + a0

Since P(k/m) = 0

P(k/m) = a3(k/m)3 + a2(k/m)2 + a1(k/m) + a0 = 0

= a3k3/m3 + a2k2/m2 + a1k/m  + a0 = 0

a0 = - (a3k3/m3 + a2k2/m2 + a1k/m)

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