Question

without actual division prove that 2 x power 4 minus x cube + 2 x square minus x + 2 is divisible by x square - 3 x + 2​

Answer 1

Answer: Not divisible

Solution:

$let \: \: p(x) = 2 {x}^{4} - {x}^{3} + 2 {x}^{2} - x + 2$
is divisible by

${x}^{2} - 3x + 2$
without actually divide ,to know that higher order polynomial is divisible by lower order

Use remainder theorem

first find the zeros of
${x}^{2} - 3x + 2 \\ \\ {x}^{2} - 2x - x + 2 = 0 \\ \\ x(x - 2) - 1(x - 2) = 0 \\ \\ (x - 2)(x - 1) = 0 \\ \\ x - 2 = 0 \\ \\ x = 2 \\ \\ x - 1 = 0 \\ \\ x = 1$
Now find P(2) and p(1)

$p(2) = 2 {(2)}^{4} - {(2)}^{3} + 2 {(2)}^{2} - (2) + 2 \\ \\ = 32 - 8 + 8 - 2 + 2 \\ \\ p(2) = 32 \\ \\ p(1) = 2 {(1)}^{4} - {(1)}^{3} + 2 {(1)}^{2} - (1) + 2 \\ \\ = 2 - 1 + 2 - 1 + 2 \\ \\ p(1) = 4$
Since in both the cases remainder is not zero,thus p(x) is not completely divisible by

${x}^{2} - 3x + 2$

Answer 2

An

Answer: Not divisible

Solution:

let \: \: p(x) = 2 {x}^{4} - {x}^{3} + 2 {x}^{2} - x + 2 \\ \\

is divisible by

{x}^{2} - 3x + 2 \\ \\

without actually divide ,to know that higher order polynomial is divisible by lower order

Use remainder theorem

first find the zeros of

{x}^{2} - 3x + 2 \\ \\ {x}^{2} - 2x - x + 2 = 0 \\ \\ x(x - 2) - 1(x - 2) = 0 \\ \\ (x - 2)(x - 1) = 0 \\ \\ x - 2 = 0 \\ \\ x = 2 \\ \\ x - 1 = 0 \\ \\ x = 1 \\ \\

Now find P(2) and p(1)

p(2) = 2 {(2)}^{4} - {(2)}^{3} + 2 {(2)}^{2} - (2) + 2 \\ \\ = 32 - 8 + 8 - 2 + 2 \\ \\ p(2) = 32 \\ \\ p(1) = 2 {(1)}^{4} - {(1)}^{3} + 2 {(1)}^{2} - (1) + 2 \\ \\ = 2 - 1 + 2 - 1 + 2 \\ \\ p(1) = 4 \\ \\

Since in both the cases remainder is not zero,thus p(x) is not completely divisible by

{x}^{2} - 3x + 2 \\