Friday
May 24, 2013

# Homework Help: Algebra

Posted by kevin on Monday, November 26, 2012 at 7:36am.

Have a couple questions was wondering if someone could check my work?

1. What are the real or imaginery solutions of the polynomial equation?
x^4-41x^2=-400

(x^2-25)(x^2-16)
x^2=25 or x=+-5
x^2=16 or x=+-4

2. Use synthetic division to find P(-3) for P(x)=x^2-2x^3-4x+4

this one I did not know how to do if someone could help

3. Find all the zeros of the eqution
x^4-6x^2-7x-6=0

messed this one up if someone could explain

4. Use Pascals Triangle to expand the binomial (d-5)^6

1d^5(-5)^0+6d^5(-5)^1+15d^4(-5)^2+20d^3(-5)^3+15d^2(-5)^4+6d^1(-5)^5+1d^0(-5)^6

d^6-30d^5+375d^4-2500d^3+9875d^2-18750d+15625

5. What is the equation of y=x^3 with the given transformation?
vertical compression by a factor of 1/7, horizontal shift 8 units to the left, reflection across the x-axis.

y=x^3
y=(1/7)x^3
y=(1/7)(x+8)^3
-y=(1/7)(x+8)^3
y=(-1/7)(x+8)^3

6. What are all the real zeroes y=(x-12)^3-7

do not understand this one either if someone wants to explain.

Thank you in advance for any help someone can give

• Algebra - kay, Monday, November 26, 2012 at 10:52am

1. Let y=x^2. Then y^2-41y+400=0 D=81>0 so real solutions y=25 or y=16 . Henece x^2=25 ie x= +- 5 OR x^2=16 hence x=+-4
2. P(-3)= (-3)^2-2(-3)^3-4(-3)+4= 9+54+12+4= 79

3. possible integer solutions:+-1,2,3,6 (dividents of constant 6)

• Algebra - Steve, Monday, November 26, 2012 at 10:54am

#1. OK

#2. Difficult to explain synthetic division without formatted text. Google it for numerous examples.

#3. x^4-6x^2-7x-6=0
Here is where synthetic division will come in handy. A little trial will show that the polynomial factors into
(x+2)(x-3)(x^2+x+1)
you can find the roots of the quadratic to get
x = -2 or 3 or -1/2 ± √3/2 i

#4. I get 9375d^2

#5. OK

#6. y=(x-12)^3-7
using difference of cubes,
y = ((x-12) - ∛7)((x-12)^2 + (x-12)∛7 + ∛49)
obviously, x = 12+∛7 is one root

The quadratic takes a little care, but solves out using the quadratic formula to

12 - 1/2 ∛7 (1+i√3)
12 + 1/2 ∛7 (-1+i√3)

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