Factoring trinomials

[mcruz65]

Find all integers b for which a^2 + ba – 50 can be factored.

Factors of – 50 = (- 50, 1), (50 – 1), (- 25, 1), (25, 1), (- 10, 1), (10, -1), (- 5,1), (5,-1), (-2,1),(2,-1)


(a – 50)(a + 1) = a2 – 50a + a – 50 = a2 – 49a – 50
(a + 50)(a – 1 ) = a2 + 50a - a + 50 = a2 + 49a – 50
(a – 25)(a + 1) = a2 + a - 25a – 25 = a2 – 24a – 25
(a + 25)(a - 1) = a2 - a + 25a – 25 = a2 + 24a – 25
(a – 10)(a + 1) = a2 + a - 10a – 10 = a2 – 9a – 10
(a + 10)(a - 1) = a2 - a + 10a – 10 = a2 + 9a – 10
(a – 5)(a + 1) = a2 + a - 5a – 5 = a2 – 4a – 5
(a + 5)(a - 1) = a2 - a + 5a – 5 = a2 + 4a – 5
(a – 2)(a + 1) = a2 + a - 2a – 2 = a2 – a – 2
(a + 2)(a - 1) = a2 - a + 2a – 2 = a2 + a – 2

Is this correct??

 

[tutor_joel]

It looks right. So your answer would be.

plus and minus 49, 24, 9, 4 and 1

 

[soroban]

Hello, mcruz65!

You've jumbled your factors . . .

Find all integers \(\displaystyle b\) for which \(\displaystyle a^2 + ba - 50\) can be factored.

. . \(\displaystyle \begin{array}{c|c|c} \text{Factors of -50} & \text{Factors} & \text{Product} \\ \hline (1,\text{-}50) & (a-1)(a+50) & a^2 + 49a - 50 \\ (\text{-}1,50) & (a+1)(a-50) & a^2 - 49a - 50 \\ (2,\text{-}25) & (a-2)(a+25) & a^2 + 23a - 50 \\ (\text{-}2,25) & (a+2)(a-25) & a^2 - 23a - 50 \\ (5,\text{-}10) & (a-5)(a+10) & a^2+5a-50 \\ (\text{-}5,10) & (a+5)(a-10) & a^2 - 5a - 50 \\ (5,\text{-}5) & (a-5)(a+5) & a^2 - 25 \end{array}\)


\(\displaystyle \text{The possible values of }b\text{ are: }\;0,\;\pm5,\;\pm23,\;\pm49\)

 

[mcruz65]

Thank you for your confirmation that I seem to know what I am doing. Can you help me with another problem?

Factor (a+4)^2-2(a+4)+1

So far this is what I have:

(a+4)(1-2+1)

a - 2a + a + 4 - 8 + 4

-2a - 8

Not sure if this is the way to do it.

 

[mmm4444bot]

Factor (a+4)^2 - 2(a+4) + 1

So far this is what I have:

(a+4)(1-2+1) This is incorrect.

If you want to factor out (a + 4), then the unsimplified factorization is:

\(\displaystyle (a + 4) \left ( (a + 4) \ - \ 2 \ + \ \frac{1}{a + 4} \right )\)

Not sure if this is the way to do it. I'm confident that this is not the factorization that your instructor wants to see.

The given expression is quadratic, in form.

For example, if we replace the expression a + 4 with the expression z, we get the following.

z^2 - 2z + 1

Can you factor that?

If so, then reverse the substitution by switching z, in your factorization, back to the expression a + 4, and simplify.

Please continue showing your work, if you want more help.

 

[mcruz65]

Thank you mmm4444bot for explaining the problem in a fashion that I can understand. Please let me know if I did it right.

(a+4)^2 - 2(a+4) + 1

[(a-4) - 1][(a-4) -1]

(a+4)^2 - (a-4) - (a-4) + 1

(a+4)^2 -2(a-4) + 1

 

[mmm4444bot]

Thank you mmm4444bot for explaining the problem in a fashion that I can understand. Please let me know if I did it right.

(a+4)^2 - 2(a+4) + 1

[(a-4) - 1][(a-4) -1]

(a+4)^2 - (a-4) - (a-4) + 1

(a+4)^2 -2(a-4) + 1 This is not your answer, is it? (heh, heh)

You've come full circle, back to the original expression.

I suggested that you reverse the substitution and simplify.

You reversed the substitution and then used FOIL to multiply it back out before simplifying.

Don't use FOIL. FOIL is for expanding; we are factoring.

You factored the quadratic polynomial z^2 - 2z + 1 as (z - 1)*(z - 1). That's good. We can write it as:

(z - 1)^2

Now, switch back to a + 4, and simplify what's inside the parentheses. You're done.

 

[mcruz65]

I sorry mmm4444bot, but I am stuck. I don't know if I have brain freeze but I can't figure this out.

[(a-4) - 1][(a-4) -1]

(a-4) -1 = 0
(a-4) = 1

:?:

 

[Mrspi]

I sorry mmm4444bot, but I am stuck. I don't know if I have brain freeze but I can't figure this out.

[(a-4) - 1][(a-4) -1]<-------where did (a - 4) come from. Didn't you originally let z = a + 4?


(a-4) -1 = 0
(a-4) = 1

:?:

You substituted z for (a + 4), right?

And your new expression was z[sup:1jv7htg3]2[/sup:1jv7htg3] - 2z + 1.....

Which you factored as (z - 1)(z - 1), or (z - 1)[sup:1jv7htg3]2[/sup:1jv7htg3]

Now, you know that z = a + 4, so you have

[(a + 4) - 1][sup:1jv7htg3]2[/sup:1jv7htg3]

Can you simplify the expression (a + 4) - 1 by combining like terms? That's all you have to do.

 

[mcruz65]

Wow! I can't believe how I allowed this problem to drive me crazy. I believe this is the correct answer. Please verify.

(a+4)^2 - 2(a+4) + 1

[(a+4)-1][(a+4)-1]

 

[mmm4444bot]

[(a + 4) - 1] * [(a + 4) - 1] This factorization is correct, but it can be simplified, so you're not done, yet.

Do you notice that both factors above are identical? When a quantity is multiplied by itself, we call that "squaring".

Here is how we write your squared product above:

[(a + 4) - 1]^2

Next, the expression inside the brackets can also be simplified because those two constants are like-terms. (The constants are 4 and -1.) In algebra, like-terms can always be combined.

Notice that, since the expression (a + 4) is not multiplied by anything, the parentheses can be removed.

[a + 4 - 1]^2

Can you finish the simplification? I mean, can you combine the like-terms 4 and -1 ?

Do that, and you're done!

 

[mmm4444bot]

I can't believe how I allowed this problem to drive me crazy.

Yeah, I'm not sure what's up with that.

I'm thinking that you need a lot more practice with the basic rules of algebra, so, if you want to succeed, try to spend as much time as you can budget working extra exercises in your math classes.

Also, you might need to slow down and concentrate more on the actual words, when reading lessons, examples, and suggestions.

For example, no less than three times did we previously tell you that the expression (a + 4) - 1 needs to be simplified.

Yet, I needed to tell you a fourth time, in my last post above. Considering why this happened might enlighten you about ways to improve your learning skills.

Cheers ~ Mark