|
Question 27805: Factor each of the following polynomials.
4s + 6st - 14st^2
Lost.
Thank you.
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website!
To factorise 4s + 6st - 14st^2
The problem should be 4s + 26st - 14st^2 or 4s + 10st - 14st^2
Consider 4s + 26st - 14st^2
=s(4+6t_14t^2)(taking s out)
= s(-14t^2 +26t +4) (rewriting so that the square in t term first,the linear term in t next and then the constant term-which is nothing but additive associativity that we have used)
=-2s(7t^2 -13t -2)----(*) (pulling out(-2) common)
Consider the quadratic expression (7t^2 -13t -2)
We have the product of the coeff of t^2 and the constant term given by
(7)X(-2) = -14 = -(1X2X7)= (-14)X(1)
(Since the product is negative and the midterm also negative, the two sets into which all the factors should form will be 0f opposite signs and to the larger numerical set we give the sign of the middle term.)
(7t^2 -13t -2)
=7t^2 +(-14t+t) -2 (writing the mid term as the sum of two terms whose product should be the product of the coeff of t^2 and the constant term )
=7t^2-14t+t -2
=(7t^2-14t)+(t -2)(by additive associativity)
= 7t(t-2) +1(t-2)
=7tp +p whee p = (t-2)
=p(7t+1)
=(t-2)(7t+1) ----(**)
Putting (**) in (*)
we have (4s + 26st - 14st^2)
= -2s(7t^2 -13t -2)----(*)
=-2s(t-2)(7t+1)
Consider 4s + 10st - 14st^2
=s(4+10t-14t^2)(taking s out)
= s(-14t^2 +10t +4) (rewriting so that the square in t term first,the linear term in t next and then the constant term-which is nothing but additive associativity that we have used)
=-2s(7t^2 -5t -2)----(*) (pulling out(-2) common)
Consider the quadratic expression (7t^2 -13t -2)
We have the product of the coeff of t^2 and the constant term given by
(7)X(-2) = -14 = -(1X2X7)= (-7)x(2)
(Since the product is negative and the midterm also negative, the two sets into which all the factors should form will be 0f opposite signs and to the larger numerical set we give the sign of the middle term.)
(7t^2 -5t -2)
=7t^2 +(-7t+2t) -2 (writing the mid term as the sum of two terms whose product should be the product of the coeff of t^2 and the constant term )
=7t^2-7t+2t -2
=(7t^2-7t)+(2t -2)(by additive associativity)
= 7t(t-1) +2(t-1)
=7tp +2p where p = (t-1)
=p(7t+2)
=(t-1)(7t+2) ----(**)
Putting (**) in (*)
we have (4s + 10st - 14st^2)
= -2s(7t^2 -5t -2)----(*)
=-2s(t-1)(7t+2)
Answer:If the given expression is (4s + 26st - 14st^2) then
(4s + 26st - 14st^2) = -2s(t-2)(7t+1)
And If the given expression is (4s + 10st - 14st^2) then
(4s + 10st - 14st^2) = -2s(t-1)(7t+2)
Note; Instead of pulling our s first and then minus 2, you may take out (-2s)
straight away.
|
|
|
| |
