Question

what is the sum of the zeroes of p(x)=x³-4x²+5x-29?​

Answer 1

4

The sum of the zeroes of $p(x)=x^{3}-4x^{2}+5x-29$ is 4.

Concept to be used:

If $\alpha_{1},\alpha_{2},\alpha_{3},...,\alpha_{n}$ be the zeroes of the polynomial

$p(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+...+a_{n}$, then

$\sum \alpha_{1}=-\dfrac{a_{1}}{a_{0}}$

From this, we can determine that for the polynomial

$p(x)=ax^{3}+bx^{2}+cx+d$

having zeroes $\alpha_{i}$, where i = 1, 2, 3, 4

then $\sum\alpha_{i}=-\dfrac{b}{a}$

Step-by-step explanation:

Here, the given polynomial is

$p(x)=x^{3}-4x^{2}+5x-29$

Comparing it with the standard form of cubic polynomial

$p(x)=ax^{3}+bx^{2}+cx+d$, we get

• a = 1, b = - 4, c = 5, d = - 29

Thus the required sum of the zeroes

= $-\dfrac{b}{a}$

= $-\dfrac{-4}{1}$

= 4

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Answer 2

The sum of the zeros of the polynomial p(x) = x³ - 4x² + 5x - 29 is 4

Given :

The polynomial p(x) = x³ - 4x² + 5x - 29

To find :

The sum of the zeros of the polynomial

Solution :

Step 1 of 4 :

Write down the given polynomial

Here the given polynomial is

p(x) = x³ - 4x² + 5x - 29

Step 2 of 4 :

Write down the coefficients

Comparing the given polynomial with general cubic polynomial p(x) = ax³ + bx² + cx + d we get

a = 1 , b = - 4 , c = 5 , d = - 29

Step 3 of 4 :

Find the relation between coefficients and zeroes

Let zeroes of the polynomial are $\displaystyle \sf{ \alpha , \beta, \gamma }$

Then we have

$\displaystyle \sf{ \alpha + \beta + \gamma = - \frac{b}{a} = - \frac{-4}{1} = 4}$

$\displaystyle \sf{ \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} = \frac{5}{1}=5 }$

$\displaystyle \sf{ \alpha \beta \gamma = - \frac{d}{a} = - \frac{-29}{1} =29}$

Step 4 of 4 :

Find sum of the zeros of the polynomial

Hence sum of the zeros of the polynomial

$\displaystyle \sf{ = \alpha + \beta + \gamma }$

$\displaystyle \sf{ = - \frac{b}{a} }$

$\displaystyle \sf{ = - \frac{-4}{1} }$

$\displaystyle \sf{ } = 4$

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