Question

the polynomial ax³-3x²+4 and 2x³-5x+a when divided by x-2 leaves remainder p and q. If p-2q = 4. find the value of a

Answer 1

x-2 = 0
   x = 2
P(x) = ax³-3x²+4
P(2) = a * 2³ - 3 * 2² + 4 = p
           a * 8 - 3 * 4 + 4 = p
                  8a - 12 + 4 = p
                          8a - 8 = p

f(x) = 2x³-5x+a = q
f(2) = 2 * 2³ - 5*2 + a = q
            2 * 8 - 10 + a = q
                16 - 10 + a = q
                         6 + a = q  

Since, p-2q = 4
             8a - 8 - 2(8+a) = 4
             8a - 8 - 16 - 2a = 4
            8a - 2a - 8 - 16 = 4
                         -6a - 24 = 4
                                -6a = 4 + 34
                                    a = 56/-6
                                    a = -9.33

Answer 2

$\huge\mathfrak{\underline{\underline{Answer :-}}}$

Let f(x) = ax³ - 3x² + 4

and

g(x) = 2x³ - 5x + a

By remainder theorem, we know that f(x) when divided by (x-2) gives a remainder equal to f(2) i.e., p = f(2) and g(x) when divided by (x-2) gives a remainder equal to g(2), i.e, = g(2).

$\sf{\underline{\underline{According\:to\:the\:given\:condition\:we\:have : }}}$

p - 2q = 4

$\implies$ $\sf{f(2)-2g(2)=4}$

$\implies$$\sf{[a(2)^3-3(2)^2+4]-2[2(2)^3-5(2)+a]=4}$

$\implies$$\sf{(8a-12+4)-2(16-10+a)=4}$

$\implies$$\sf{(8a-8)-2(a+6)=4}$

$\implies$$\sf{6a-20=4}$

$\implies$$\sf{6a=24}$

$\implies$a=$\sf\frac{24}{6}$

$\implies$ a = $\bf{4}$

Hence, the value of a is $\bf{4}$