Question

2y³+y²-2y-1
pless tell me​

Answer 1

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Question:-

• 2y³+y²-2y-1

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Solution:-

If we will put y =1

so ,

$⇢2 {y}^{3} + {y}^{2} + 2y - 1 \\ \\⇢2 {(1)}^{3} + {1}^{2} - 2(1) - 1 \\ \\ ⇢2 + 1 - 2 - 1 \\ \\ ⇢3 - 3 \\ \\ ⇢0$

So y = 1 is the factor of equation

See attachment

After this we get

$2 {y}^{2} + 3y + 1$

Now if we factorise it

$⇢2 {y}^{2} + 3y + 1 \\ \\ ⇢2 {y}^{2} + (2 + 1)y + 1 \\ \\ ⇢2 {y}^{2} + 2y + y + 1 \\ \\ ⇢y(2y + 1) + 1(2y + 1) \\ \\ ⇢(2y + 1)(y + 1)$

So factors of this are (y-1), ( y +1) and (2y +1)

If we see 2y + 1

2 y+1 =0

y =- 1/2

If we see y + 1

y +1 =0

y =-1

If we see y -1

y -1 = 0

y =1

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Answer 2

$\huge\bf{ \underline{Question}}$

• $\sf{2\ y^{3}+\ y^{2}-2y-1}$

$\huge\bf{ \underline{Answer}}$

• $\sf{±1\:,\:\dfrac{-1}{2}}$

$\huge\bf{ \underline{Solution}}$

${}$

$\mapsto\sf{2\ y^{3}+\ y^{2}-2y-1\:=\:0}$

$\sf{}$

$\mapsto\sf{\ y^{2}(2y+1)-1(2y+1)\:=\:0}$

${}$

$\mapsto\sf{(2y+1)(\ y^{2}-1)\:=\:0}$

$\sf{Now\:,\:find\:value \:of\:y}$

• $\sf{2y+1\:=\:0}$

$\sf{ \purple{ \underline{ \boxed{y\:=\:\dfrac{-1}{2}}}}}$

• $\sf{\ y^{2}-1\:=\:0}$

$\sf{ \purple{ \underline{ \boxed{y\:=\:±1}}}}$

${}$