Question

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Factoring Trinomials

Answer 1

Question 1: \[9x ^{2}+9x-10\]
A:\[(3x-5)(3x+2)\]
B: \[(3x+5)(3x-2)\]
C: \[(3x-1)(3x+10)\]
D: Not Factorable

Question 2: \[10x ^{2}+19x+6\]
A: \[(2x+5)(3+2x)\]
B: \[(5x+2)(2x+3)\]
C: \[(2x-5)(3-2x)\]
D: Not Factorable

Answer 2

Question 3: \[18x ^{2}-54x+112\]
A:\[(9x+2)(2x-56)\]
B:\[(9x-2)(2x+56)\]
C:\[(9x+2)(2x+56)\]
D:Not Factorable

Answer 3

Question 1 is B.

Answer 4

No answer dear @BioEpic, Against Open Study rules

Answer 5

Question 4: \[3x^{2}-x-10\]
A:\[(3x-2)(x+5)\]
B:\[(3x+2)(x-5)\]
C:\[(x-2)(3x+5)\]
D: Not Factorable

Question 5: \[8x ^{2}+22x+5\]
A:\[(2x-5)(4x-1)\]
B:\[(2x+5)(4x+1)\]
C:\[(2x+5)(4x-1)\]
D: Not Factorable

Question 6: \[14x ^{2}-13x+5\]
A:\[(7x-1)(2x+5)\]
B:\[(7x-5)(2x+1)\]
C:\[(-7x-1)(2x+5)\]
D: Not Factorable

Answer 6

to factor a Trinomial if it factorable
think about \((a+b)(a+c)=a^2+ac+ab+bc=a^2+a(b+c)+bc\)
so think about two numbers that multiply to give you \(bc\) and add up to give you \(b+c\)

Answer 7

hey! one question at a time!
for more question you have to open new thread. However your question are all same kind
so one will be enough, the others you have to do them yourself after seeing how to do it

Answer 8

let's go with Q1
\(9x ^{2}+9x-10\)
this trinomial if it is foctorable it would look this way
\(9x ^{2}+9x-10=(9x~~~~~~~ )(x~~~~~~)\)
the empty space if for the missing numbers and signs
where did i get 9x and x you might ask!
it is because the leading coefficient is 9 and the power 2 so it would be x times x
9x times x gives \(9x^2\)
so for that reason we chose that

Answer 9

Are you following?

Answer 10

yeh

Answer 11

to know what are the numbers we are missing
we ask what are the two numbers that multiply to give us -10
at the same time add up to 9
to answer such a question we look at factors of 10
{1,2,5,10}
if we pick 2 and 5 they won't work since they don't add up to 9 in any case
1 and 10 is the right choice then!
\(-1\times10=-10\)
\(-1+10=9\)
if i choose -10 and 1
i would get \(-10\times1=-10\)
\(-10+1=-9\)
but we are not looking for -9
we are looking for 9
so -1 and 10 are the correct choice
then we go back to our parenthesis
\(9x ^{2}+9x-10=(9x\color{red}{-1})(x+\color{red}{10})\)
this way it is wrong since we weill get 90x-x =89x not 9x
so the right order is
\(9x ^{2}+9x-10=(9x-\color{blue}{10})(x+\color{blue}{-1})\)

Answer 12

opps mistake
\(9x ^{2}+9x-10=(9x+\color{blue}{10})(x\color{blue}{-1})\)
i meant to put this way the last line

Answer 13

you could do \((3x ~~~~)(3x~~~~)\)
since 3x times 3x is 9x^2

Answer 14

eh hold on my bad I'm not thinking straight!
let's go with (3x )(3x )
case
2 and 5 multiply to 10
as i said above! if i have -2 and 5 the middle term will be
-6x+15x=9x
if it is 2 and -5 we will get for the middle term
6x-15x=-9x
so -2 and 5 are good
it would be \((3x+5)(3x-2)\)

Answer 15

what i did first in incorrect! i forgot that we are multiplying by 9x
so the middle term would be 10x-9x=x
but we want 9x not x so not a good choice

Answer 16

ok, thanks you've helped a lot

Answer 17

welcome