show that 2x+1 is a factor of the polynomial 2x³+x²-6x-3.hence factorize the polynomial.
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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)
marina1:
are you sure that the answer is right
yes.
you can check with the examples given in your book.
Answered by
1
2x3 + x2 -6x-3 can be factorise as (2x +1)(x^2-3)
Hence it can be prived that (2x+1) is the factor of polynomial
your answer is too short .please do this sum in detail.
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