Question

Using factor theorem factorise it...
Please do it. it's urgent...​

Answer 1

2y^3+y^2-2y-1

=y^2(2y+1)-1(2y+1)

=(2y+1)(y^2-1)

Using a^2-b^2=(a+b)(a-b)

=(2y+1)(y+1)(y-1)

Answer 2

Question :-

Using the factor theorem Factorise 2y³ + y²- 2y - 1

Answer :-

Given :-

Polynomial :- 2y³ + y² - 2y - 1

Required to find :-

• Factorise the given expression ?

Condition mentioned :-

• Factor theorem

Solution :-

Given polynomial :-

2y³ + y² - 2y - 1

Consider the given polynomial as ;

p ( y ) = 2y³ + y² - 2y - 1

Using the factor theorem let's find the factors for the given p ( y ) .

( using hit trial and error method )

Let assume that y - 1 is the factor

So,

=> y - 1 = 0

=> y = 1

p ( 1 ) =

2 ( 1 )³ + ( 1 )³ - 2 ( 1 ) - 1 = 0

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

2 + 1 - 2 - 1 = 0

0 = 0

LHS = RHS

Hence,

y - 1 is the factor of p ( y )

Now, we need to perform long division and in result we need to Factorise the quotient .

So,

$\rm y - 1 \big)2 {y}^{3} + {y}^{2} - 2y - 1 \big(2 {y}^{2} + 3y + 1 \\ \: \: \rm \: \: \: \: \: \: \: \: \: \: \: \: 2 {y}^{3} - 2 {y}^{2} \\ \: \: \: \: \: \: \: \: \: \: \underline{ \rm ( - ) \: \: \: ( + ) \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm 3 {y}^{2} - 2y \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: 3 {y}^{2} - 3y \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \underline{( - ) \: \: ( + ) \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: 1y - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: \: \: 1y - 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{( - ) \: \: ( + ) \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline { \: \: \: \: \: \: \: \: \: \large{0} \: \: \: \: \: \: \: \: \: \: \: }$

Hence,

Quotient = 2y² + 3y + 1

Now we need to Factorise this quotient

2y² + 3y + 1

Sum = 2y + 1y

product = 2 x 1

So,

➜ 2y² + 3y + 1

➜ 2y² + 2y + 1y + 1

➜ 2y ( y + 1 ) + 1 ( y + 1 )

➜ ( y + 1 ) ( 2y + 1 )

Therefore ,

2y³ + y² - 2y - 1 is factorised into ( y - 1 ) , ( y + 1 ) & ( 2y + 1 )