Question

if one zero of the polynomial x^2-13x+40 is 5, what is the other zero

Answer 1

Given polynomial x^2-13x+40 One zero is 5 We know that sum of zeros = -b/a Here a= 1,b= -13,c= 40 Let other zero be p: P+5=-( -13)/1 P= 13-5= 8 Answer will be 8

Answer 2

The other zero is 8.

Given:

• A quadratic polynomial
• ${x}^{2} - 13x + 40$
• One zero is 5.

To find:

• Find the other zero of polynomial.

Solution:

Concept to be used:

Relationship between zeroes of polynomial and coefficients of polynomial:

The standard quadratic equation is

$\bf a {x}^{2} + bx + c, \: a \ne0$

if $\alpha \: and \: \beta$ are zeros of polynomial.

Then relationship between zeroes of polynomial and coefficients of polynomial is given by

$\alpha + \beta = \frac{ - b}{a} ...eq1$

and

$\alpha \beta = \frac{c}{a} ...eq2$

Step 1:

Write the coefficients of given polynomial.

Here polynomial is

${x}^{2} - 13x + 40$

so,

$a = 1$

$b = - 13$

and

$c = 40$

Step 2:

Let the given zero is $\alpha$

So,

$\alpha = 5$

Let the other zero is $\beta$

Step 3:

Put the values in eq1

$5 + \beta = \frac{ - ( - 13)}{1}$

or

$5 + \beta = 13$

or

$\beta = 13 - 5$

or

$\red{\beta = 8}$

Thus,

The other zero is 8.

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