Question

if x=1 and x=-2 are the roots of the polynomials x^3+4x^2-3ax+b then find the value of a and b I need The process​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\rm \: x = 1,2 \: are \: roots \: of \: {x}^{3} + {4x}^{2} - 3ax + b$

Let assume that

$\rm :\longmapsto\:f(x) \: = \: {x}^{3} + {4x}^{2} - 3ax + b - - (1)$

Now,

It is given that 1 is the root of f(x).

We know,

Factor theorem states that if x = a is the root of f(x) then f(a) = 0.

So, using factor theorem, we have

$\rm :\longmapsto\:f(1) = 0$

$\rm :\longmapsto\: {1}^{3} + {4(1)}^{2} - 3a(1) + b = 0$

$\rm :\longmapsto\:1 + 4 - 3a + b = 0$

$\rm :\longmapsto\:5- 3a + b = 0$

$\bf\implies \:3a - b = 5 - - - (2)$

Also,

It is given that x = 2 is a root of f(x)

So, By Using factor theorem,

$\rm :\longmapsto\:f(2) = 0$

$\rm :\longmapsto\: {(2)}^{3} + {4(2)}^{2} - 3a(2) + b = 0$

$\rm :\longmapsto\:8 + 16 - 6a + b = 0$

$\rm :\longmapsto\:24 - 6a + b = 0$

$\bf\implies \:6a - b = 24 - - - (3)$

On Subtracting equation (2) from equation (3), we get

$\rm :\longmapsto\:3a = 19$

$\bf\implies \:a = \dfrac{19}{3}$

On substituting the value of a, in equation (2), we get

$\rm :\longmapsto\:3 \times \dfrac{19}{3} - b = 5$

$\rm :\longmapsto\:19 - b = 5$

$\rm :\longmapsto\: - b = 5 - 19$

$\rm :\longmapsto\: - b = - 14$

$\bf\implies \:b = 14$

Additional Information :-

Remainder Theorem:-

This theorem states that if a polynomial f (x) is divided by linear polynomial x - a, then remainder is f(a).