# Show that 2x+1 is a factor of the polynomial 2x³+x²-6x-3.hence factorize the polynomial.

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by marina1

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by marina1

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If (2x+1) is a factor of (2x³+x²-6x-3), then zero of (2x+1) will also be a zero of (2x³+x²-6x-3).

zero of (2x+1) can be found by equating it to 0.

2x+1 = 0

⇒2x = -1

⇒x = -1/2

Put x=-1/2 in (2x³+x²-6x-3). we get

(2x³+x²-6x-3)

= 2(-1/2)³ + (-1/2)² - 6(-1/2) - 3

= 2(-1/8) + (1/4) - 6(-1/2) - 3

= -2/8 + 1/4 + 6/2 - 3

= -1/4 + 1/4 + 3 - 3

= 0

Since -1/2 is also a zero of (2x³+x²-6x-3), (2x+1) is a factor of (2x³+x²-6x-3).

Factorizing:

2x³ + x² - 6x - 3

= (2x³ + x²) - (6x + 3)

=x²(2x+1) - 3(2x+1)

= (x²-3)(2x+1)

x²-3 can be further factorized as (x-√3)(x+√3)

So 2x³+x²-6x-3 = (x-√3)(x+√3)(2x+1)

zero of (2x+1) can be found by equating it to 0.

2x+1 = 0

⇒2x = -1

⇒x = -1/2

Put x=-1/2 in (2x³+x²-6x-3). we get

(2x³+x²-6x-3)

= 2(-1/2)³ + (-1/2)² - 6(-1/2) - 3

= 2(-1/8) + (1/4) - 6(-1/2) - 3

= -2/8 + 1/4 + 6/2 - 3

= -1/4 + 1/4 + 3 - 3

= 0

Since -1/2 is also a zero of (2x³+x²-6x-3), (2x+1) is a factor of (2x³+x²-6x-3).

Factorizing:

2x³ + x² - 6x - 3

= (2x³ + x²) - (6x + 3)

=x²(2x+1) - 3(2x+1)

= (x²-3)(2x+1)

x²-3 can be further factorized as (x-√3)(x+√3)

So 2x³+x²-6x-3 = (x-√3)(x+√3)(2x+1)