Question

Find the values of a&b so that 2x^3+ax^2+x+b has factor x+2& 2x-1

Answer 1

Given f(x) = 2x^3 + ax^2 + x + b.

Given that 2x - 1 and x + 2 are the factors of f(x).

= > 2x - 1 = 0

= > x = 1/2

Plug x = 1/2 in f(x), we get

= > 2(1/2)^3 + a(1/2)^2 + (1/2) + b = 0 

= > 1/4 + a/4 + 1/2 + b = 0

= > 3/4 + a/4 + b = 0

= > a + 3 + 4b = 0   

= > a + 4b = -3 ------ (1)




When x + 2:

= > x + 2 = 0

= > x = -2.

Plug x = -2 in f(x), we get

 = > 2(-2)^3 + (a)(-2)^2 + (-2) + b = 0

= > -16 + 4a - 2 + b = 0

= > 4a + b - 18 = 0 

= > 4a + b = 18  ------ (2)


On solving (1) * 4 & (2), we get

4a + 16b = -12

4a + b = 18

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

         15b = -30

            b = -30/15.

            b = -2



Substitute b = -2in (1), we get

= > a + 4b = -3

= > a + 4(-2) = -3

= > a - 8 = -3

= > a = -3 + 8

a = 5.



Therefore the value of a = 5 and b = -2.


Hope this helps!

Answer 2

Answer: The value of a and b  is 71 and 2

Explanation- Given equation is $2x^2+ax^2+x+b$, $2x-1$ and $x+2$ is the factor of given function.

Find- The value of a and b

Solution- According to the question

$2x-1=0\\x=\frac{1}{2} \\x+2=0\\x=-2$

Now $x=\frac{1}{2}$ put in f(x)

$2x^3+ax^2+x+b=0$

$2\times \frac{1}{8}+a\times \frac{1}{4}+\frac{1}{2}+b=0$

$\frac{1}{4}+\frac{a}{4} + \frac{1}{2}+b=0$

1 and 2  $\frac{1+a+2+4b}{4}=0$

$a+4b+3=0$

$a+4b=-3....(1)$

Now put $x=-2$ in f(x)

$2x^3+ax^2+x+b=0$

$2(-2)^3+a(-2)^2+(-2)+b=0\\-16+4a-2+b=0\\4a+b=18......(2)$

Now solve equations 1 and 2

$a+4b=-3 \times 4\\4a+b=18$

$4a+16=-12\\4a+b=18$

$-15b=-30$

$b=2$

Put the value of b in equation 1

$a+4b=-3\\a+4\times 2=-3\\a=-11$

Hence, the value of a and b is -11 and 2

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