Find the third zero of the cubic polynomial ax3+bx2+cx + d if two of the zeroes are each equal to 0

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If two of the zeros of the cubic polynomial ax3+bx2+cx+d are 0 then the third zero is a b/a b b/a c c/a d d/a

Solution

Let the required roots be p ,q ,r let the two roots let p and q be 0 (given)

we know that$\to$ sum of roots = -b/a

$\to$ p + q + r = -b/a
$\to$ 0 + 0 + r = -b/a
$\to$ r = -(b/a)
so option a is correct

Mathematics

Secondary School Mathematics X

Standard X

Let `alpha = 0, beta=0` and y be the zeros of the polynomial

f(x)= ax3 + bx2 + cx + d 

Therefore

`alpha + ß + y= (-text{coefficient of }X^2)/(text{coefficient of } x^3)`

`= -(b/a)`

`alpha+beta+y = -b/a`

`0+0+y = -b/a`

`y = - b/a`

`\text{The value of}  y - b/a`

Hence, the correct choice is `(c).`

Page 2

If two zeros x3 + x2 − 5x − 5 are \[\sqrt{5}\ \text{and} - \sqrt{5}\], then its third zero is

Let `alpha = sqrt5` and `beta= -sqrt5` be the given zeros and y  be the third zero of x3 + x2 − 5x − 5 = 0 then

By using `alpha +beta + y = (-text{coefficient of }x^2)/(text{coefficient of } x^3)`

`alpha + beta + y = (+(+1))/1`

`alpha + beta + y = -1`

By substituting `alpha = sqrt5` and `beta= -sqrt5` in `alpha +beta+y = -1`

`cancel(sqrt5) - cancel(sqrt5) + y = -1`

` y = -1`

Hence, the correct choice is`(b)`

Concept: Relationship Between Zeroes and Coefficients of a Polynomial

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Page 3

Given `alpha, beta,y` be the zeros of the polynomial x3 + 4x2 + x − 6

Product of the zeros  = `(\text{Constant term })/(\text{Coefficient of}\x^3) = (-(-6))/1 =6`

The value of Product of the zeros is 6.

Hence, the correct choice is `( c ).`