Question

713.44
Using the remainder theorem, factorise the
expression 2x3 + x2 – 2x - 1 completely.
(a) (x - 1)(x-4) (2x + 1)​

Answer 1

Answer:

(x-1)(x+1)(2x+1)

Step-by-step explanation:

we can write any expression a(x) as

a(x) = b(x) q(x) + r

Let us denote the zero of the linear polynomial b(x) by k. This means that b(k) = 0. If we plug in x as k in the starred relation above, we have a(k) = b(k) q(k) + r

When a polynomial a(x) is divided by a linear polynomial b(x) whose zero is x equal to k, the remainder is given by r=a(k).

so, we can solve it as

$2 x^{2} + x^{2} - 2x -1$ =0 for x=1

so, we can divide  $2 x^{2} + x^{2} - 2x -1$ by x-1 and it gives $2x^{2} +3x+1$ as result without remainder.

$2x^{2} +3x+1$ =0 for x= -1

so, we can now divide equation $2x^{2} +3x+1$ by x+1 and it gives 2x+1

so, the factors of given equation is (x-1)(x+1)(2x+1)

Answer 2

Given :   2x³  + x² - 2x - 1

To Find : Using the remainder theorem, factorise

Solution:

2x³  + x² - 2x - 1

polynomial p(x) divided by (x - a)

Then p(a) is the remainder

if p(a) = 0 then (x - a) is  a factor

p(x)  = 2x³  + x² - 2x - 1

p(1) = 2(1) + 1² - 2(1) - 1  = 0

Hence x - 1 is one of the factor

p(-1)  = 2(-1) + (-1)² - 2(-1) - 1  = - 2 + 1 + 2 - 1 = 0

Hence ( x- (-1)) = x + 1 is also a factor

2x³  + x² - 2x - 1   is divisible by (x + 1)(x - 1)

(x + 1)(x - 1) = x² - 1

              2x + 1

x² - 1   _|  2x³  + x² - 2x - 1   |_

               2x³         -2x

             _______________

                         x²        - 1

                         x²        - 1

             _______________

                              0

2x³  + x² - 2x - 1  =  (x + 1)(x - 1)(2x + 1)

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