Question

Find the values of a and b so that the polynomial (x^4+ax^3-7x^2-8x+b) is exactly divisible by (x+2) as well as (X+3)​

Answer 1

Solution =》

$p(x) = x {}^{4} + ax {}^{3} - 7x {}^{2} - 8x + b = 0 \\ g(x) = x + 2 \\ = x + 2 = 0 \\ = x = - 2$

Putting x=-2

$( - 2 ){}^{4} + a( - 2) {}^{3} - 7 \times ( - 2) {}^{2} - 8 \times ( - 2) + b = 0 \\ = 16 + ( - 8)a - 7 \times 4 + 16 + b = 0 \\ = 16 - 8a - 28 + 16 + b = 0 \\ = 4 - 8a + b = 0 \\ = - 8a + b = - 4............(1)$

Putting x=-3

$x {}^{4} + ax {}^{3} - 7x {}^{2} - 8x + b = 0 \\ = ( - 3) {}^{4} + a( - 3) {}^{3} - 7( - 3){}^{2} - 8 \times ( - 3) + b = 0 \\ = 81 - 27a - 63 + 24 + b = 0 \\ = 42 - 27a + b = 0 \\ - 27a + b = - 42........(2)$

On subtracting eq 2 from eq 1

we get ,

a=2

Putting a=2in eq 1

-8a+b=-4

-8×(2)+b=-4

-16+b=-4

b=-4+16

b=12

So the value of a=2 and b=12

Answer 2

Question

Find the values of a and b so that the polynomial (x^4+ax^3-7x^2-8x+b) is exactly divisible by (x+2) as well as (X+3)

Solution

Given :-

• Polynomial , x⁴ + ax³ - 7x² - 8x + b = 0
• Factor are (x+2) & (x+3)

Find :-

• Value of a & b

Explanation

We know, If (x+2) & (x+3) are factor of this equation ,

So, Value of x = -2 & -3 satisfies of this equation.

So, Keep first value of x = -2

==> (-2)⁴ + a(-2)³ - 7(-2)² - 8(-2) + b = 0

==> 16 -8a - 28 + 16 +b = 0

==> -8a + b = 28 - 16 - 16

==> 8a - b = -28 + 32

==>8a - b = 4 ---------------equ(1)

Again, Keep x = -3

==> (-3)⁴ + a(-3)³ - 7(-3)² - 8(-3) + b = 0

==> 81 - 27a - 63 + 24 + b = 0

==> -27a + b = -105+63

==> 27a - b = 42 ----------------equ(2)

Subtract equ(1) & equ(2)

==> 8a - 27a = 4 - 42

==>-19a = -38

==> a = 38/19

==>a = 2

Now, Keep value of a in equ(1)

==> 8 * 2 - b = 4

==> b = 16 - 4

==> b = 12

Hence

• Value of a = 2
• Value of b = 12

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