Question

$\huge\pink{\mid{\underline{\overline{\tt HELLO}}\mid}}$

GENIUS...SOLVE THIS

For what value of a and b are the zeroes of
$q(x) = {x}^{3} + 2 {x}^{2} + a$
are also the zeroes of the polynomial
$p(x) = {x}^{5} - {x}^{4} - 4 {x}^{3} + 3 {x}^{2} + 3x + b$
⁉⁉⁉⁉???


(Also answer my last English question)


All the best!!!⭕⭕⭕
​

Answer 1

Solution

Given that the zeroes of q(x)= x^3 +2x^2+a are also the zeroes of the polynomial p(x)= x^5-x^4-4x^3+3x^2+ 3x+b. q(x) is a factor of p(x). Then, we use a division algorithm.

$= { {x}^{5} - {x}^{4} - 4 {x}^{3} + 3 {x}^{2} + 3x +b} \div {x}^{3} + 2 {x}^{2} + a \\ = - (1 + a) {x}^{2} + (3 + 3a)x + (b - 2a)$

If (x^3+ 2x^2+a) is a factor of (x^5- x^4- 4x^3+ 3x^2+ 3x+ b), then remainder should be zero.

==) -(1+a)x^2+(3+3a)x+(b-2a)= 0

==) 0.x^2+ 0.x+0

On comparing the coefficient of x, we get

= a+1= 0

= a= -1

and b-2a= 0

= b=2a

= b= 2(-1) = -2-----------{value of a= -1}

HOPE IT HELP

Answer 2

= { {x}^{5} - {x}^{4} - 4 {x}^{3} + 3 {x}^{2} + 3x +b} \div {x}^{3} + 2 {x}^{2} + a \\ = - (1 + a) {x}^{2} + (3 + 3a)x + (b - 2a)=x

5

−x

4

−4x

3

+3x

2

+3x+b÷x

3

+2x

2

+a

=−(1+a)x

2

+(3+3a)x+(b−2a)

If (x^3+ 2x^2+a) is a factor of (x^5- x^4- 4x^3+ 3x^2+ 3x+ b), then remainder should be zero.

-(1+a)x^2+(3+3a)x+(b-2a)= 0

0.x^2+ 0.x+0

On comparing the coefficient of x, we get

= a+1= 0

= a= -1

and b-2a= 0

= b=2a

= b= 2(-1) = -2££££-{value of a= -1