Question

if 1 and -3 are zeros of the polynomial p(x)= x^3- ax^2 - 13x+b,find values of a and b.

Answer 1

Step-by-step explanation:

Given -

1 and -3 are zeroes of polynomial

p(x) = x³ - ax² - 13x + b

To Find -

Value of a and b.

Now,

p(1) = (1)³ - a(1)² -13(1) + b

= 1 - a - 13 + b = 0

• = - a + b = 12 ........... (a)

And

p(-3) = (-3)³ - a(-3)² -13(-3) + b

= - 27 - 9a + 39 + b = 0

= - 9a + b = -12 .......... (b)

Subtracting (a) and (b)

- a + b = 12

- 9a + b = - 12

(+) (-) (+)

---------------------------

8a = 24

a = 3

Substituting the value of a = 3 on (a)

= - a + b = 12

= - 3 + b = 12

= b = 12 + 3

= b = 15

Hence,

The value of a = 3 and b = 15

Verification -

The Polynomial is x³ - 3x² - 13x + 15

= x³ - x² - 2x² + 2x - 15x + 15

= x²(x - 1) - 2x(x - 1) - 15(x - 1)

= (x - 1) (x² - 2x - 15)

= (x - 1) (x² + 3x - 5x - 15)

= (x - 1) [x(x + 3) - 5(x + 3)]

= (x - 1) (x - 5) (x + 3)

Zeroes are -

x - 1 = 0 and x - 5 = 0 and x + 3 = 0

x = 1 , 5 , -3

Hence,

Our Answer is Absolutely Correct.

Bcoz The zeroes of the polynomial comes same as given.

Verified..

Answer 2

Given,

$p(x) = {x}^{3} - a {x}^{2} - 13x + b$

and also (1) and (-3) are zero's of the polynomial p(x)

Zero of the polynomial :

A real number a is a zero of a polynomial p(x), if p(a) = 0.

• Now by using this theorem we can say that p(1) and p(-3) are equal to zero.

$\implies \: p(1) = {1}^{3} - a {x}^{2} - 13x + b$

$\implies \: p(1) = 1 - a - 13 + b$

$\implies p(1) = - 12 - a + b$

$\implies \: 0 = - 12 - a + b$

$\implies \: a - b = ( - 12) :eq(1)$

similarly,

we need to find the value of p(-3)

$\implies \: p( - 3) = {( - 3)}^{3} - a {x}^{2} - 13x + b$

$\implies \: p( - 3) = (- 27) - 9a + 39 + b$

$\implies \: p( - 3) = - 9a + b + 12$

$\implies 0 = ( - 9a) + b + 12$

$\implies 9a - b = 12 : eq(2)$

Now, multiply eq(1) with (-9)

$\implies \: \: ( - 9a) + 9b = (12 \times 9)$

$\implies \: 9a - b = 12$

Now, by solving these equations , we get

$8b = 12(9 + 1)$

$8b = 120$

$b = 15$

Now, by substituting the value of b in eq(1) we get,

$a - 15 = ( - 12)$

$a = 15 - 12$

$a = 3$

Therefore, the values of a and b are 3 , 15 respectively.