Question

If the zeros of the polynomial (x+1)(x+9)+8 are a and b then the zeros of the polynomial (x+a)(x+b)-8 are: (A) 1 and 9 (B) -4 and -6 (C) 4 and 6 (D) Cannot be determined​

Answer 1

Answer:

Option b is the right answer

Step-by-step explanation:

Answer 2

The zeroes of the polynomial (x+a)(x+b)-8 are 1 and 9.

Option (a) is the correct option.

Given,

A polynomial (x+1)(x+b)+8 and its zeroes are a and b.

To find,

The zeroes of (x+a)(x+b)-8

Solution,

First, let's find the zeroes of the given polynomial.

(x+1)(x+b)+8

${x}^{2} + 10x + 17$

To find zeroes,

$x = ( - B + \sqrt{ {B}^{2} - 4AC } ) \div 2A$

and

$x = (- B - \sqrt{ {B}^{2} - 4AC }) \div 2A$

So, the zeroes of the polynomial by solving are,

-5+√2 and -5-√2.

Put them in another polynomial, (x+a)(x+b)-8

(x-5+2√2)(x-5-2√2)-8

${x}^{2} - 5x - 2 \sqrt{2} x -5x + 25 + 10 \sqrt{2} + 2 \sqrt{2} - 10 \sqrt{2x - 8 - 8}$

${x}^{2} - 10x + 9$

Let's find the zeroes,

(x-9)(x-1)=0

x=1 and x=9.

So, zeroes are 1and9.

Hence, option (a) is the correct answer.

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