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Question Express the polynomial h(x)=x4−6x3+38x2−62x+29h(x) = x^{4} - 6x^{3} + 38x^{2} - 62x + 29 as a product of linear factors, given that x=1x = 1 is a zero of multiplicity two.

Studdy Solution

STEP 1

Assumptions
1. The polynomial is h(x)=x4−6x3+38x2−62x+29h(x)=x^{4}-6x^{3}+38x^{2}-62x+29.
2. The polynomial has a zero at x=1x=1 with multiplicity two.

STEP 2

Since 11 is a zero of multiplicity two, (x−1)2(x-1)^2 is a factor of h(x)h(x).

STEP 3

We will divide h(x)h(x) by (x−1)2(x-1)^2 to find the other factors.

STEP 4

Perform the division using polynomial long division or synthetic division.

STEP 5

Set up the synthetic division with 11 as the root to be used for division.

STEP 6

Carry out the synthetic division process:
1−638−62291↓1−533−291−533−290\begin{array}{r|rrrrr} & 1 & -6 & 38 & -62 & 29 \\ 1 & \downarrow & 1 & -5 & 33 & -29 \\ \hline & 1 & -5 & 33 & -29 & 0 \\ \end{array}

STEP 7

The result of the synthetic division gives us the coefficients of the quotient polynomial:
q(x)=x2−5x+33q(x) = x^{2} - 5x + 33

STEP 8

Now we need to factor the quadratic polynomial q(x)q(x), if possible.

STEP 9

Calculate the discriminant of the quadratic polynomial q(x)q(x):
Δ=b2−4ac\Delta = b^2 - 4ac

STEP 10

Plug in the values for aa, bb, and cc from q(x)q(x):
Δ=(−5)2−4(1)(33)\Delta = (-5)^2 - 4(1)(33)

STEP 11

Calculate the discriminant:
Δ=25−132=−107\Delta = 25 - 132 = -107

STEP 12

Since the discriminant is negative, q(x)q(x) does not have real roots and hence cannot be factored over the real numbers.

STEP 13

We can express q(x)q(x) as a product of two complex conjugate linear factors.

STEP 14

Find the complex roots of q(x)q(x) using the quadratic formula:
x=−b±Δ2ax = \frac{-b \pm \sqrt{\Delta}}{2a}

STEP 15

Plug in the values for aa, bb, and Δ\Delta:
x=5±−1072x = \frac{5 \pm \sqrt{-107}}{2}

STEP 16

Express the complex roots in terms of ii where i2=−1i^2 = -1:
x=5±i1072x = \frac{5 \pm i\sqrt{107}}{2}

STEP 17

Now we have the two complex roots, which are complex conjugates of each other:
x=5+i1072andx=5−i1072x = \frac{5 + i\sqrt{107}}{2} \quad \text{and} \quad x = \frac{5 - i\sqrt{107}}{2}

STEP 18

Express these roots as linear factors:
(x−5+i1072)(x−5−i1072)(x - \frac{5 + i\sqrt{107}}{2})(x - \frac{5 - i\sqrt{107}}{2})

STEP 19

Multiply out the factors to get them in the standard form (x−p)(x−q)(x - p)(x - q):
(x−52−i1072)(x−52+i1072)(x - \frac{5}{2} - \frac{i\sqrt{107}}{2})(x - \frac{5}{2} + \frac{i\sqrt{107}}{2})

STEP 20

Now we have all the factors of h(x)h(x):
h(x)=(x−1)2(x−52−i1072)(x−52+i1072)h(x) = (x-1)^2(x - \frac{5}{2} - \frac{i\sqrt{107}}{2})(x - \frac{5}{2} + \frac{i\sqrt{107}}{2})

STEP 21

Finally, express h(x)h(x) as a product of linear factors:
h(x)=(x−1)2(x−(52+i1072))(x−(52−i1072))h(x) = (x-1)^2 \left(x - \left(\frac{5}{2} + \frac{i\sqrt{107}}{2}\right)\right) \left(x - \left(\frac{5}{2} - \frac{i\sqrt{107}}{2}\right)\right)
This is the expression of h(x)h(x) as a product of linear factors, including the given zero of multiplicity two and the two complex conjugate roots.

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