Question

Find the remainder of the polynomial
(1)
P(x) = 2x^3- x^2- 4x +8 is divided by X+1
(2)
p(X)=2x^3-3x^2+4x-5 I'd divided by 2x+1
(3)
p(X)=X^3-ax^2+7x+8 Is divided by x-1 leaves remainder 7.Find 'a'​

Answer 1

Step-by-step explanation:

$1) \: \: \: \: x + 1 = 0 \\ \: \: \: \: \: \large{ \boxed{x = \underline{ \underline{- 1}}}} \\ \\ \: \: \: \: p(x) = 2 {x}^{3} - {x}^{2} - 4x + 8 \\ \: \: \: \: = 2 {( - 1)}^{3} - {( - 1)}^{2} - 4( - 1) + 8 \\ \: \: \: \: = (2 \times - 1) - (1) - (4 \times - 1) + 8 \\ \: \: \: \: = - 2 - 1 + 4 + 8 \\ \: \: \: \: \large{ \boxed{= \: \underline{ \underline{9}}}} \\ \\ 2) \: \: \: \: 2x + 1 = 0 \\ \: \: \: \: 2x = - 1 \\ \: \: \: \: \: \large{ \boxed{x = \underline{ \underline{\frac{ - 1}{2} }}}} \\ \\ \: \: \: \: p(x) = 2 {x}^{3} - 3 {x}^{2} + 4x - 5 \\ \: \: \: \: = 2 {( \frac{ - 1}{2}) }^{3} - 3 { (\frac{ - 1}{2}) }^{2} + 4( \frac{ - 1}{2} ) - 5 \\ \: \: \: \: = (2 \times \frac{ - 1}{8} ) - (3 \times \frac{ - 1}{4} ) + (4 \times \frac{ - 1}{2} ) - 5 \\ \: \: \: \: = \frac{ - 1}{4} - \frac{ - 3}{4} + ( - 2) - 5 \\ \: \: \: \: = ( \frac{ - 1 - ( - 3)}{4} ) - 2 - 5 \\ \: \: \: \: = ( \frac{ - 1 + 3}{4} ) - 7 \\ \: \: \: \: = \frac{2}{4} - 7 \\ \: \: \: \: = \frac{1}{2} - 7 \\ \: \: \: \: = \frac{1 -14}{2} \\ \: \: \: \: \large{ \boxed{ = \underline{ \underline{ \frac{ - 13}{2}}}}} \\ \\ 3) \: \: \: \: x - 1 = 0 \\ \: \: \: \: \: \large{ \boxed{x = \underline{ \underline{1}}}} \\ \\ \: \: \: \: p(x) = {x}^{3} - a {x}^{2} + 7x + 8 = 7 \\ \: \: \: \: \: {1}^{3} - a {(1)}^{2} + 7(1) + 8 = 7 \\ \: \: \: \: \: 1 - (a \times 1) + 7 + 8 = 7 \\ \: \: \: \: \: 1 - a + 15 = 7 \\ \: \: \: \: \: 16 - a = 7 \\ 16 - 7 = a \\ \large{ \boxed{\underline{ \underline{9}} = a}}$