Question

If the polynomials ax3+4x2+3x-4 and x3 -4x+a leave the same  remainder when divided by (x-3), find the value of 'a'.
plz don't give unnecessary answers ​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x - 3 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{x = 3}$

$p(x) = a {x}^{3} + 4{x}^{2} + 3x - 4$

$p(x) = {x}^{3} - 4x + a$

$as \: both \: eqations \: leave \: same \: remainder \\ when \: divided \: by \: (x - 3) \: so$

$a {x}^{3} + 4{x}^{2} + 3x - 4 = {x}^{3} - 4x + a$

$a(3) {}^{3} + 4 {(3)}^{2} + 3(3) - 4 = {3}^{3} - 4(3) + a$

$27a + 36 + 9 - 4 = 27 - 12 + a$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 27a + 41 = 15 + a$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 27a - a = 15 - 41$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 26a = - 26$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \red{a = - 1}}$

$\mathcal\orange{Please\:mark\:as\:brainliest.....}$