Question

If x=-1\3 is a zero of the polynomial p(x) =27x cube - ax square - x+3 then find the value of a

Answer 1

Heya !!
Here's your answer..⬇
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x = -1/3

p(x) = 27x³ - ax² - x + 3

p( -1/3 ) = 27( -1/3 )³ - a( -1/3 )² - ( -1/3 ) + 3

=> 27( -1/27 ) - a( 1/9 ) + 1/3 + 3 = 0

=> -27/27 - a/9 + 1/3 + 3 = 0

=> -1 - a/9 + 1/3 + 3 = 0

=> (-1 × 9 - a + 3 + 27)/9 = 0

=> - 9 - a + 3 + 27 = 0

=> a = 30 - 9

=> a = 21
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Hope it helps..
Thanks :)

Answer 2

Value of a is 21.

Given:

• $p(x) = 27 {x}^{3} - a {x}^{2} - x + 3$
• $x = \frac{ - 1}{3}$ is a zero of p(x).

To find:

• Find the value of x.

Solution:

Concept/Theorem to be used:

• If x=a is a zero of polynomial, then (x-a) is its factor.
• Apply factor theorem.
• Factor theorem: It states that if (x-a) is factor of polynomial p(x), then p(a)=0.

Step 1:

Put the value of zero in polynomial.

$p\left( - \frac{1}{3} \right) =0$

or

$27 {\left( - \frac{1}{3} \right)}^{3} - a {\left( - \frac{1}{3}\right )}^{2} -\left ( - \frac{1}{3} \right)+ 3 = 0$

Step 2:

Simplify and find the value of x.

$-\frac{27}{ 27} - \frac{a}{9} + \frac{1}{3} + 3 = 0$

or

$- 1 - \frac{a}{9} + \frac{1}{3} + 3 = 0$

or

$2 - \frac{a}{9} + \frac{1}{3} = 0$

or

$\frac{18 - a + 3}{9} = 0$

or

$- a + 21 = 0$

or

$- a = - 21$

or

$\bf \red{a = 21}$

Thus,

Value of a is 21.

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