Question

When a polynomial p(x) = x
⁴ – 2x³ + 3x² – ax + b is divisible by x – 1 and x + 1, the

remainders are 5 and 19 respectively. Find the remainder when p(x) is divided by x – 2.​

Answer 1

$\large\sf\underline{Given}$

When p(x) is divided by x-1 and x+1 the remainder are 5 and 19 respectively.

$\tt\pink{∴\:p(1)=5}$

$\sf\: x⁴ -2x³ + 3x² - ax + b=5$

$\sf⟹\: 1⁴ - 2 \times 1³ + 3 \times 1² - a \times 1 + b=5$

$\sf⟹\: 1 -2 + 3 - a + b=5$

$\sf⟹\: -1 + 3 - a + b=5$

$\sf⟹\: 2 - a + b=5$

$\sf⟹\: - a + b=5-2$

$\tt\purple{\: - a + b=3-------(i)}$

$\tt\pink{∴\:p(-1)=19}$

$\sf\: x⁴ - 2x³ + 3x² - ax + b=19$

$\sf⟹\:(- 1)⁴ - 2 \times(- 1)³ + 3 \times (-1) ² – a \times(- 1) + b=19$

$\sf⟹\: 1 - 2 \times (-1) + 3 \times 1 + a + b=19$

$\sf⟹\: 1 + 2+3 + a + b=19$

$\sf⟹\: 3+3 + a + b=19$

$\sf⟹\: 6+ a + b=19$

$\sf⟹\: a + b=19-6$

$\tt\purple{\: a + b=13--------(ii)}$

Adding the two equations :

$\sf⇒\:( - a + b ) + ( a + b ) = 3 + 13$

$\sf⇒\:-a + b + a + b = 3 + 13$

$\sf⇒\:2b=16$

$\sf⇒\:b=\frac{16}{2}$

$\tt\red{\:⇒b=8}$

Putting b = 8 and -a + b = 3 in equation (i) , we get

$\tt\purple{\: - a + b=3}$

$\sf⇒\: -a + 8 = 3$

$\sf⇒\:-a=3-8$

$\sf⇒\:-a=-5$

$\tt\red{\:⇒a=5}$

Putting the values of a and b in

$\sf\:p(x) = x⁴ – 2x³ + 3x² – 5x + 8$

The remainder when p(x) is divided by (x-2) is equal to p(2).

So, Remainder

$\sf\:p(2) = x⁴ - 2x³ + 3x² - 5x + 8$

$\sf⟹\:p(2) = 2⁴ - 2 \times 2³ + 3 \times 2² - 5 \times 2 + 8$

$\sf⟹\:p(2) = 16 - 2 \times 8 + 3 \times 4 - 5 \times 2 + 8$

$\sf⟹\:p(2) = 16 - 16 + 12 - 10 + 8$

$\sf⟹\:p(2) =12 - 10 + 8$

$\sf⟹\:p(2) =2 + 8$

$\tt\pink{⟹\:p(2) =10}$

Thnku ❣

Answer 2

check the attachment....