solve and show all work for a2 – a – 20 and a2 – 5a – 20
what are you solving?
Factor each trinomial below. Please show your work and check your answer.
ok for the first one we have a^2-a-20 a^2-1a-20 so first step is to ask yourself what two integers multiply to by -20 and add up to be -1?
What do you mean?
\[ax^2+bx+c \\ \text{ when } a=1 \text{ and if the expression is "factorable" } \\ \text{ then you should be able to find } \\ \text{ two integers so that when you multiply them together you get } c \\ \text{ and when you add them together you get } b\] for example: \[x^2+5x+6 \\ \text{ the question here first to answer is: } \\ \text{ what two integers multiply to be 6 } \\ \text{ and add up to be 5}\] well 2(3)=6 and 2+3=5 so the factored form of \[x^2+5x+6 \text{ is } (x+2)(x+3)\]
a2 – a – 20 (a )(a ) -4*5 10*-2 (a+4)(a-5) a2 -5a+4a-2 -a a2 – 5a – 20 (a )(a ) 10*-2 -5*4 (a-5)(a+4) a2 + 12ab + 27b2 (a )(a ) 9*3 (a+3b)(a+9b) a2 +9ab+3ab+27b a2 +12ab+27b
Right?
yes to the first one: a^2-a-20=(a-5)(a+4) since -5*4=-20 and -5+4=-1 no to the second one: a^2-5a-20 this does not equal (a-5)(a+4) -5(4)=-20 but -5+4 isn't -5
and the third one looks good
a2 – a – 20 (a )(a ) -4*5 10*-2 (a+4)(a-5)
so you got 2/3 correct
you just need to fix the second one
Ok
I will try
a^2-5a-20 are there two integers that multiply to be -20 and add up to be -5?
possible pairings that multiply to be -20: 1(-20) -1(20) 2(-10) -2(10) 4(-5) -4(5)
do any of those pairs also add up to -5?
which equation is true if any: 1+(-20)=-5 -1+20=-5 2+(-10)=-5 -2+10=-5 4+(-5)=-5 -4+5=-5
None
you are right
They are all false
so that means that this is a prime trinomial
means it is not "factorable"
I put factorable in quotation marks because it only really means it isn't factorable over the integers
Ok
So I am not sure on the second one
it isn't factorable
it is prime
Ok thanks
a^2-5a-20 does not factor
unless you can factor over the reals?
which I don't think that is the instructions
No but how would I write this just say it is prime?
you can either say not factorable or prime
which ever vocabulary your teacher prefers really but either saying is fine in general
Part 1: Factor each trinomial below. Please show your work and check your answer. For the first one: x^2 - 8 x + 15 x^2 - 5x - 3x + 15 x(x-5) -3(x-5) (x-5)(x-3) The check: x(x-3)-5(x-3) x^2 – 3x-5x+15 x^2-8x+15 Question 2: a2 – 5a – 20 Prime and cannot be factored. Question: 3 a^2 - a - 20 a^2 -5a+4a-20 a(a-5)+4(a-5) (a-5)(a+4) Question 4: a^2 + 12ab + 27b^2 a^2 + 9ab + 3ab + 27b^2 a(a+9b) + 3b(a+9b) (a+9b)(a+3b) Question 5: 2a2 + 30a + 100 2(a^2 + 15a + 50) the values that add to give 15a multiply to 50a^2 are 10a and 5a 2(a^2 + 5a + 10a + 50) 2[a(a+5) +10(a+5)] 2(a+5)(a+10) Question 6: Part 2: (5 points) It’s your turn to be a game show host! As you know, in the game of Math Time, the contestants are given an answer and they must come up with the question that corresponds to the given answer. Your task for this portion of the assignment is to create two different “answers” (and the questions that accompany them) that the host could use for the final round of Math Time. The questions and answers you create must be unique. Check out the example and hint below, if needed. 1 point for creating the two questions and answers. 2 points per question/answer for accuracy. Example: Question for Host: x2 + 6x + 5 is the product of these two binomials. Expected Question from Contestant: What is (x + 5)(x + 1)? HINT: Create the two binomials first and then use the distribution method to find their simplified product. The simplified product is the answer and the two binomial factors, the expected question. Hosts question 1: The product of these two binomials is x2+x-20 Answer: What is (x+5)(x-4) Hosts question 2: The product of these two binomials is x2+7xy+12y Answer: What is (x+3y)(x+4y)
how does this look?
@myininaya
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