X 0 and x 1 arethe roots of the polynomial f x 2x 3 3x 2 ax b find the value of a and b

Text Solution

Solution : Given: `x=0` and `x=-1` `f(x)=2x^3-3x^2+ax+b` Putting `x=0` in `f(x)`, we get `f(0)=2(0)^3-3(0)^2+a(0)+b` `b=0` Putting `x=-1` and `b=0` in `f(x)`, we get `f(-1)=2(-1)^3-3(-1)^2+a(-1)+0` `0=-2-3-a` `a=-5` Hence, the value of `a=-5` and `b=0`.

If x = 0 and x = −1 are the roots of the polynomial f(x) =2x3 − 3x2 + ax + b, find the value of a and b.

The given polynomial is

f(x) =2x3 − 3x2 + ax + b

If x=0 is zeros of the polynomial f(x), then f(0) = 0

`2 xx (0)^3 -3 xx (0)^2 + a xx 0 +b = 0`

`0-0 +0 +b = 0`

`b = 0 ` ....... (1)

Similarly, if x = − 1 is the zeros of the polynomial of f(x),

Then, f (-1) = 0

`2x (-1)^3 - 3 xx (-1)^2 + a xx ( - 1)+ b = 0 `

` - 2 -3-a +b =0 `

` -5-a +b = 0`

Putting the value of b from equation (1)

` -5 -a + 0 =0`

`a= -5`

Thus,

`a=-5`

` b=0`

Concept: Factorising the Quadratic Polynomial (Trinomial) of the type ax2 + bx + c, a ≠ 0.

Is there an error in this question or solution?

Page 2

The given polynomial is

`f (x) = x^3 + 6x^2 + 11x + 6`

Here, f(x) is a polynomial with integer coefficient and the coefficient of highest degree term is 1. So, the integer roots of f(x) are factors of 6. Which are ±1, ±2, ±3, ±6 by observing.

`f(-1) = (-1)^3 + 6xx (-1)^2 + 11(-1) + 6`

` = -1 + 6 - 11 + 6`

` = -12 + 12`
= 0

Also,

`f(-2) = (-2)^3 + 6(-2)^2 + 11(-2) + 6`

` = -8 + 6 xx 4 - 22 + 6`

` = -8 + 42 - 22 + 6`

`= 30 - 30`

` = 0`

And similarly,

f(−3) = 0

Therefore, the integer roots of the polynomial f(x) are −1, −2, − 3.

Page 3

The given polynomial is

`f(x) = 2x^3 + x^2 - 7x - 6`f(x) is a cubic polynomial with integer coefficients. If \[\frac{b}{c}\] is rational root in lowest terms, then the values of b are limited to the factors of 6 which are \[\pm 1, \pm 2, \pm 3, \pm 6\] and the values of c are limited to the factor of 2 as \[\pm 1, \pm 2\] Hence, the possible

rational roots are \[\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}\].

Since, `f(2) = 2.2^3 + 2^2 - 7.2 - 6 = 0`

So, 2 is a root of the polynomial`f(x) = 2x^3 + x^2 - 7x - 6`

Now, the polynomial can be written as,

`f(x) = (x-2)(2x^2 + 5x + 3)`

Also,

`f(-1) = (-1-2) (2 - 5 + 3) = 0`

Therefore,

`f(x) = (x - 2) (x+ 1) (2x + 3)`

Hence, the rational roots of the polynomial `f(x) = 2x^3 + x^2 - 7x - 6` are 2, – 3/2 and – 1.

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