Question

Page
the
Using
the remainder theorem, factorise
2x3 + 3x²-9x-10​

Answer 1

Answer:

write it properly

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Answer 2

Answer:

(x-2)(x+1)(2x+5)

Step-by-step explanation:

Let p(x) = 2x³ + 3x² - 9x - 10

To factorize this polynomial, we have to check the factors of 10 i.e. ±1, ±2, ±5. It is found by calculation that p(2) = 0 as follows

p(2) = 2*2³ + 3*2² - 9*2 - 10

= 16 + 12 - 18 - 10

= 28 - 28

= 0

Hence by remainder theorem (x - 2) is a factor.

Now let us divide p(x) by (x-2)

x-2 ) 2x³ + 3x² - 9x - 10 ( 2x²+ 7x+ 5

2x³ - 4x²

⁽⁻⁾ ⁽⁺⁾

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

7x² - 9x

7x² - 14x

⁽⁻⁾ ⁽⁺⁾

-------------------

5x - 10

5x - 10

------------

So, p(x) = (x-2)(2x² + 7x + 5)

The quadratic expression can be factorized by splitting middle term.

p(x) = (x-2)( 2x² + 2x + 5x + 5)

= (x-2)[2x(x+1) + 5(x+1)]

= (x-2)(x+1)(2x+5)