Question

Is 2t+1 a factor of 4t³+4t²-t-1​

Answer 1

$\color{aqua}answer : -$

Yes ,( 2t + 1 )is a factor of (4t³+4t²-t-1)

$\color{violet}explanation : -$

☆ To find :

● Is 2t +1 is a factor of 4t³+4t²-t-1

☆Assuming:

●Let p(x) = 4t³+4t²-t-1 and g(x)= 2t+1

☆Solution:

●g(x)=0

$- > 2t + 1 = 0 \\ = > 2t = - 1 \\ = > t = \frac{ - 1}{2}$

●g(x) will be a factor of p(x) only,if 4t³+4t²-t-1 is divided by 2t+1 leaving remainder zero.

●By the remainder theorem ,we know that when p(x) is divided by ( 2t+1) then the remainder is p (-1/2)

$now \: p( \frac{ - 1}{2} ) =4 {( \frac{ - 1}{2}) }^{3} + 4 {( \frac{ - 1}{2} )}^{2} - (\frac{ - 1}{2}) - 1 \\ = 4 \times \frac{ - 1}{8} + 4 \times \frac{1}{4} - (\frac{ - 1}{2} ) - 1 \\ = \frac{ - 1}{2} + 1 - ( \frac{ - 1}{2} ) - 1 \\ = \frac{ - 1 + 1 + 1 - 1}{2} = \frac{0}{2} = 0$

●Thus,when p(x) is divided by g(x) ,the remainder is zero.

●Therefore,( 2t + 1 )is a factor of (4t³+4t²-t-1)

hope it helps... :)