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lara [203]
3 years ago
13

1. Write the following polynomial in factored form. Show your work.

Mathematics
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

1. y = x^2( x^-3) (x+2)

2. x=0 multiplicity 2  and x=4  multiplicity 2

3.  y  = 2x^3 -x^2 -50x +25

4.y =x^3 -6x^2 +5x+12

Step-by-step explanation:

1.  y = x^4 - x^3 - 6x^2

Factor out an x^2

y = x^2( x^2 -x-6)

We can factor the polynomial inside the parenthesis

What multiplies to -6 and add to -1

-3*2 = -6  and -3+2 =-1

y = x^2( x^-3) (x+2)


2.  f(x) = x^4 - 8x^3 + 16x^2

We need to factor out an x^3

f(x) = x^2(x^2 - 8x + 16)

We should notice inside the parentheses is the difference of two cubes

(a^2 -2ab-b^2) = (a-b)^2  where a=x   and b=-4

f(x) = x^2(x-4)^2

We can use the zero product property to find the roots

0 =  x^2(x-4)^2

x^2 = 0

x= 0 with multiplicity 2  ( because it is a square)

(x-4)^2 =0

x-4 =0

x=4  with multiplicity 2  ( because it is a square)

3.  y = (x - 5)(x + 5)(2x - 1)

We know (x-5) (x+5) is the difference of squares

(a-b) (a+b) = a^2 -b^2   where a=x  and b=5

y = (x ^2- 5^2 )(2x - 1)

y = (x ^2- 25 )(2x - 1)

Now we FOIL  first outer inner last

y = x^2 * 2x + -1*x^2 + -2x*25 -1*-25

y  = 2x^3 -x^2 -50x +25

Standard form is from highest exponent to lowest exponent.


4.  Using the zero product property we know that

(x-a)(x-b)(x-c) =0  where a,b,c  are the zero's of the function

(x--1) (x-3) (x-4) = 0

(x+1)(x-3) (x-4)

Now FOIL the first two terms

(x^2 -3x+1x-3)(x-4)

(x^2 -2x-3) (x-4)

Now we multiply x by the first term then multiply -4 by all the terms in the  first term

x^2 *x -2x^2 -3x -4x^2 +8x+12

Combine like terms

x^3 -6x^2 +5x+12

This is now in standard form, since the exponents go from highest to lowest power

y =x^3 -6x^2 +5x+12

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