Given that two of the zeroes of the cubic polynomial ax3 bx2 cx + d are 0 the third zero is

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

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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).`

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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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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 ).`

Text Solution

`(-b)/a` `b/a` `c/a` `(-d)/a`

Answer : A

Solution : Let ` alpha, 0, 0` be the zeros of `ax^(3) + bx^(2) + cx + d.` Then, <br> sum of zeros = `(-b)/a rArr alpha+0+0=(-b)/a rArr alpha = (-b)/a.` <br> Hence, the third zeros is `(-b)/a.`

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Given, the cubic polynomial is ax³ + bx² + cx + d.

Two zeros of the polynomial are zero.

We have to find the third zero of the polynomial.

Let first zero be 𝛼, so 𝛼 = 0

Let second zero be ꞵ, so ꞵ = 0

We know that, if 𝛼, ꞵ and 𝛾 are the zeroes of a cubic polynomial ax³ + bx² + cx + d, then

Sum of the roots is 𝛼 + ꞵ + 𝛾 = -b/a

By the property, 0 + 0 + 𝛾 = -b/a

Therefore, the third zero is -b/a.

✦ Try This: Given that two of the zeroes of the cubic polynomial rx³ + sx² + tx + u are 0, the third zero is

Given, the cubic polynomial is rx³ + sx² + tx + u

Two zeros of the polynomial are zero

We have to find the third zero of the polynomial.

Let first zero be 𝛼, so 𝛼 = 0

Let second zero be ꞵ, so ꞵ = 0

We know that, if 𝛼, ꞵ and 𝛾 are the zeroes of a cubic polynomial ax³ + bx² + cx + d, then

Sum of the roots is 𝛼 + ꞵ + 𝛾 = -b/a

Here, a = r and b = s

By the property, 0 + 0 + 𝛾 = -b/a

𝛾 = -s/r

Therefore, the third zero is -s/r

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 2

NCERT Exemplar Class 10 Maths Exercise 2.1 Solved Problem 2

Summary:

Given that the two zeros of the cubic polynomial ax³ + bx² + cx + d are zero, the third zero is -b/a

☛ Related Questions:

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• If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then a = -7, b = -1, a = . . . .
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