Question

For the polynomial below, 1 is a zero of multiplicity two.

f(x)=x² - 6x³ +62x² - 110x+53
Express f(x) as a product of linear factors.

Answer 1

Given polynomial: (f(x) = x^2 - 6x^3 + 62x^2 - 110x + 53)

Divisor (repeated factor corresponding to the zero (x = 1)): (x - 1)

Step 1: Perform Long Division

-6x + 68

_______________________

x - 1 | x^2 - 6x^3 + 62x^2 - 110x + 53

- (x^2 - x)

________________

-5x^2 - 109x

+ 5x^2 - 5x

______________

-105x + 53

+ 105x - 105

_______________

158

Step 2: Express (f(x)) as a Product of Factors

[ f(x) = (x - 1)^2 * (x^2 - x - 105) ]

Factor the Quadratic Inside

To factor the quadratic(x^2 - x - 105), we need to find two numbers whose product is ((-105)) and whose sum is ((-1)) (coefficient of (x)). These numbers are (-15) and (14).

[ f(x) = (x - 1)^2 * (x - 15)(x + 7) ]

So, the polynomial (f(x)) can be expressed as a product of linear factors:

[ f(x) = (x - 1)^2 * (x - 15)(x + 7) ]

Answer 2

`f(x)=(x-1)^2(x-2+7\:i)(x-2-7\:i)`

Explanation:

Given polynomial:

`f(x)=x^4-6x^3+62x^2-110x+53`

Multiplicity refers to the number of times a particular linear factor is repeated in the factorization of the polynomial. Therefore, if 1 is a zero of multiplicity two, one of the factors of the polynomial f(x) is (x - 1)².

Expand (x - 1)²:

`(x-1)^2=(x-1)(x-1)=x^2-2x+1`

Now, divide f(x) by the expansion of (x - 1)² using long division:

`\large \begin{array}{r}x^2-4x+53\phantom{)}\\x^2-2x+1{\overline{\smash{\big)}\,x^4-6x^3+62x^2-110x+53\phantom{)}}\\{-~\phantom{(}\underline{(x^4-2x^3+x^2)\phantom{-bwwwww))))}}\\-4x^3+61x^2-110x+53\phantom{)}\\-~\phantom{()}\underline{(-4x^3+8x^2-4x)\phantom{)))))))))}}\\53x^2-106x+53\phantom{)}\\-~\phantom{()}\underline{(53x^2-106x+53)}\\0\phantom{)}\end{array}`

Therefore, the expression for f(x) as a product of factors would be:

`f(x)=(x-1)^2(x^2-4x+53)`

As we have been asked to express f(x) as a product of linear factor, we need to factor the quadratic factor using the quadratic formula.

`\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=(-b \pm √(b^2-4ac))/(2a)\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}`

In the case of x² - 4x + 53:

• a = 1
• b = -4
• c = 53

Therefore:

`x=(-(-4) \pm √((-4)^2-4(1)(53)))/(2(1))`

`x=(4 \pm √(-196))/(2)`

`x=(4 \pm √(14^2 \cdot (-1)))/(2)`

`x=(4 \pm √(14^2) √(-1))/(2)`

`x=(4 \pm 14 \;i)/(2)`

`x=2 \pm 7 \;i`

So, the quadratic factor can be expressed as:

`(x^2+4x+53)=(x-(2-7\;i))(x-(2+7\;i))`

`(x^2+4x+53)=(x-2+7\:i)(x-2-7\:i)`

Therefore, polynomial f(x) as a product of linear factors is:

`f(x)=(x-1)^2(x-2+7\:i)(x-2-7\:i)`