SOLUTION: For what values of t can 10x^2+tx+8 be written as the product of two binomials? I don't really get what we're supposed to do in this question. Please explain.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: For what values of t can 10x^2+tx+8 be written as the product of two binomials? I don't really get what we're supposed to do in this question. Please explain.      Log On


   

Question 1174668: For what values of t can 10x^2+tx+8 be written as the product of two binomials?
I don't really get what we're supposed to do in this question. Please explain.
Found 4 solutions by josgarithmetic, ikleyn, MathLover1, greenestamps:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
You can try several factorization attempts to find which one will work. 
(10x  )(x   )
(10x  8)(x   1)

The x terms can be 18x.


highlight_green%28%2810x%2B8%29%28x%2B1%29=10x%5E2%2Bhighlight%2818%29x%2B8%29

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.

            According to the context, the coefficients of linear binomials are (should be) integer numbers.

            Otherwise,  the problem has  INFINITELY  MANY  solutions. 


Consider 

    10x^2 + tx + 8 = (ax+b)*(cx+d).


Then we want to have  ac = 10,  bd = 8.


We have these cases


    a= 10, c= 1,  b= 8, d= 1;   (10x+8)*(1x+1) = 10x^2 + 18x + 8;   t = 18.

    a=  5, c= 2,  b= 8, d= 1;   ( 5x+8)*(2x+1) = 10x^2 + 21x + 8;   t = 21.



    a= -10, c= -1,  b= 8, d= 1;   (-10x+8)*(-1x+1) = 10x^2 - 18x + 8;  t = -18.

    a=  -5, c= -2,  b= 8, d= 1;   ( -5x+8)*(-2x+1) = 10x^2 + 21x + 8;  t = -21.


It is only the beginning of an analysis.

The full number of cases is  MUCH  MORE.

I showed the idea to you.

Having a patience,  you may continue and complete it  ON  YOUR  OWN.


            REMEMBER:  you must satisfy two equalities   ac= 10,   bd= 8  in integer numbers by trial and error;

            try all appropriate positive and negative integer pairs.  Then for  "t"  you have this values   t = ad + bc.


I wish you pleasant pastime  (!)


=============

Ignore the post by  @josgarithmetic,  since it is  IRRELEVANT  to the posed problem.


Do not distract yourself reading the post from @MathLover1.


Learn from good sources (!) (!)



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

10x%5E2%2Btx%2B8+
compare to ax%5E2%2Bbx%2Bc, =>+a=10, b=t, and c=8
you need
ac=10%2A8=80
factors of 80 are:±1 | ±2 | ±4+| ±5 | ±8+| ±10 | ±16+| ±20 | ±40 | ±80+
following sums are potential values for t
1%2B80=81
2%2B40=42
4%2B20=24
5%2B16=21
8%2B10=18
same is with negative values

so t= ±18,±21,±24,±42,±81
check some:
t= ±18
if t=+18%7D%7D+we+have%0D%0A%0D%0A%7B%7B%7B10x%5E2%2B18x%2B8 ....we can write 18x as 10x+%2B8x
10x%5E2%2B10x+%2B8x%2B8...group
%2810x%5E2%2B10x%29+%2B%288x%2B8%29
10x%28x%2B1%29+%2B8%28x%2B1%29
%2810x+%2B8%29%28x%2B1%29
if t=-18
10x%5E2-18x%2B8%7B%7B%7B+....we+can+write+%7B%7B%7B-18x+as -10x+-8x
10x%5E2-10x+-8x%2B8...group
%2810x%5E2-10x%29+-%288x-8%29
10x%28x-1%29+-8%28x-1%29
%2810x+-8%29%28x-1%29
t= ±21
if t=21+which can be written as 16%2B5
10x%5E2%2B21x%2B+8
10x%5E2%2B16x%2B5x%2B+8
%2810x%5E2%2B16x%29%2B%285x%2B+8%29
2x%285x%2B8%29%2B%285x%2B+8%29
%282x+%2B+1%29+%285x+%2B+8%29+
do same using t=-21, and you will get %282+x+-+1%29+%285+x+-+8%29+


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


10x%5E2%2Btx%2B8+=+%28ax%2Bb%29%28cx%2Bd%29

Since the leading coefficient and constant term are both positive, the coefficients of a and c can be either both positive or both negative; likewise for b and d.

But allowing both pairs to be either positive or negative doesn't provide any different linear terms in the product. For example...

%282x%2B3%29%283x-1%29+=+6x%5E2%2B7x-3 and %28-2x-3%29%28-3x%2B1%29+=+6x%5E2%2B7x-3

So, for determining the number of possible values of the linear coefficient t, we can assume coefficients a and c are positive. Then the possible factorizations with integer coefficients have...
(1) (a,c) =(10,1) or (5,2) or (2,5) or (1,10); and
(2) (b,d) = (8,1) or (4,2) or (2,4) or (1,8) OR (-8,-1) or (-4,-2) or (-2,-4) or (-1,-8)

Every combination of choices from (1) and (2) will produce a different value for t.

If the question asked only for the NUMBER of different possible values for t, we would know the answer at this point: 4 choices for the pair of leading coefficients and 8 choices for the pair of constants, so 4*8=32 different values of t.

However, the question asks for a list of all the possible values of t; so there is still some work for you to do to finish the problem.

Note that in making the complete list of the possible values of t, you can use only the positive pairs of values for the constants; whatever (positive) value of t you get, the corresponding negative value of t can be obtained using the corresponding pair of negative constants.

So determine the t values using the positive choices for constants b and d to get 16 of the 32 possible values for t. Then the remaining 16 possible values of t are the corresponding negative values.