# verifying answers for polynomials

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• Jan 13th 2010, 11:06 AM
jay1
verifying answers for polynomials
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
• Jan 13th 2010, 12:10 PM
bigwave
latex
Quote:

Originally Posted by jay1
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!

hope you learn to use laTex
this is my understanding of your equation

$\displaystyle \frac{1}{4a^2} + \frac{3}{4a} = 1$

$\displaystyle \frac{1}{4a^2} + \frac{3}{4a}\times\frac{a}{a} \Rightarrow \frac{1+3a}{4a^2} = 1$

$\displaystyle 4a^2 - 3a -1 = 0$

$\displaystyle (4a-1)(a+1)= 0$

$\displaystyle a= -1$

$\displaystyle a=-\frac{1}{4}$
• Jan 13th 2010, 12:17 PM
jay1
Is there a place that will show me how to use this "laTex"? BigWave, was my answer correct? Thanks for your help.
• Jan 13th 2010, 12:23 PM
abender
Quote:

Originally Posted by jay1
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!

If problem 1 is what I interpreted it to be, which is this:

$\displaystyle \frac{1}{4a^2} + \frac{3}{4}a = 1$

then

$\displaystyle a=1, a=\frac{1+\sqrt{13}}{6}, \text{ and } a=\frac{1-\sqrt{13}}{6}$.

However, if problem 1 is this:

$\displaystyle \frac{1}{4a^2} + \frac{3}{4a} = 1$

then

$\displaystyle a=1, a= -\frac{1}{4}$.

And if problem 1 is this:

$\displaystyle \frac{1}{4}a^2 + \frac{3}{4}a = 1$

then

$\displaystyle a=1, a= -4$.
• Jan 13th 2010, 12:35 PM
Soroban
Hello, jay1!

Quote:

1) Solve: .$\displaystyle \tfrac{1}{4}a^2 + \tfrac{3}{4}a \:=\:1$

My answer: .$\displaystyle a=4,\;a=1$ . . . . no

We have: .$\displaystyle \tfrac{1}{4}a^2 + \tfrac{3}{4}a \:=\:1$

Multiply by 4: .$\displaystyle a^2 + 3a \:=\:4 \quad\Rightarrow\quad a^2 + 3a - 4 \:=\:0$

Factor: .$\displaystyle (a-1)(a+4) \:=\:0$

Therefore: .$\displaystyle a\;=\;1,\:-4$

Quote:

2) Simplify: .$\displaystyle 2ab^4 -3a^2b^2 - ab^4 + a^2b^2$

I have: .$\displaystyle ab^4- 2a^2b^2$ . . . . Yes!

Quote:

3) Simplify: .$\displaystyle -2(4y^2 +3z^3 +5) + 3(2y^2 -5z^3 +3)$

My answer: .$\displaystyle -21z^3 - 2y^2 -1$ . . . . Right!

• Jan 13th 2010, 12:36 PM
abender
PROBLEM 2:

$\displaystyle 2ab^4 - 3 a^2b^2 - ab^4 + a^2b^2 = ab^4 -2a^2b^2$

You are correct.
• Jan 13th 2010, 12:41 PM
abender
PROBLEM 3:

$\displaystyle -2(4y2+3z3+5) + 3(2y2-5z3+3) = (-8y^2 - 6z^3 - 10) + (6y^2 - 15z^3 + 9)$ $\displaystyle = -8y^2 - 6z^3 - 10 + 6y^2 - 15z^3 + 9 = -21z^3 - 2y^2 - 1$

Good job again. Problem 3 is correct.

-Andy
• Jan 13th 2010, 12:59 PM
bigwave
mystery equation
who had the mystery equation correct....
just curious..
• Jan 13th 2010, 01:08 PM
jay1
Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
• Jan 13th 2010, 01:10 PM
jay1
BigWave, I think that it was Soroban. Thanks again!
• Jan 13th 2010, 03:21 PM
abender
Quote:

Originally Posted by jay1
Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?

$\displaystyle x^2 + 4x - 45 = 0$

$\displaystyle (x+9)(x-5) = 0$

$\displaystyle x = 5, -9$

So, reverse the signs for your answer to this problem.
• Jan 13th 2010, 03:27 PM
abender
Quote:

Originally Posted by jay1
2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 [COLOR=black]I have ab4 - 2a2b2

$\displaystyle 2ab^4 - 3a^2b^2 -ab^4 + a^2b^2 = ab^4 - 2a^2b^2$

Atta boy (assuming you are a guy).
• Jan 13th 2010, 03:35 PM
abender
3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?

$\displaystyle y^2 + 12y + 35 = (y+7)(y+5) \implies y=-5,-7$

You can always double-check your answers when factoring quadratics by "FOIL"ing.

Good luck.

-Andy
• Jan 13th 2010, 06:46 PM
jay1
Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?
• Jan 13th 2010, 07:34 PM
abender
Quote:

Originally Posted by jay1
Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?

No.

$\displaystyle (5a^3 + 3a -2) - (4a^3 + a^2 + 5) = 5a^3 + 3a - 2 - 4a^3 - a^2 - 5 = a^3 - a^2 + 3a - 7$

The minus before the parenthesis changes the sign of each term inside the parentheses.

-Andy
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