Question

It's Urgent Please

The polynomial ax^3+3x^2-3 and 2x^2-5x+a when divided by (x-4) leave the remainder R1 and R2.find the value of a in this case. R1+R2=0

Answer 1

To find the value of a, polynomial ax^3+3x^2-3 and 2x^2-5x+a when divided by (x-4) leave the remainder R1 and R2.find the value of a in this case. R1+R2=0

$x - 4)a {x}^{3} + 3 {x}^{2} - 3(a {x}^{2} + (3 + 4a)x + 12 + 16a \\ \: \: \: \: \: \: \: \: \: \: \: \: a {x}^{3} - 4a {x}^{2} \\ \: \: \: \: \: - - - - - - - - \\ \: \: \: \: \: \: \: \: \: (3 + 4a) {x}^{2} - 3 \\ \: \: \: \: \: \: \: \: \: (3 + 4a) {x}^{2} - 4(3 + 4a)x \\ - - - - - - - - - - \\ \: \: \: \: \: 12x + 16ax - 3 \\ \: \: \: \: \: 12x + 16ax - 48 - 64a \\ - - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: 45 + 64a = R1$
$x - 4)2 {x}^{2} - 5x + a(2x + 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: 2 {x}^{2} - 8x \\ \: \: \: \: \: - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3x + a \\ \: \: \: \: \: \: \: \: \: \: \: \: \: 3x - 12 \\ \: \: \: \: \: \: \: \: \: \: \: - - - - - \\ \: \: \: \: \: \: \: \: 12 + a = R2$
$R1 + R2 = 0 \\ \\ 45 + 64a + 12 + a = 0 \\ \\ 57 + 65a = 0 \\ \\ a = \frac{ - 57}{65}$