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Standard X

Mathematics

Zeroes of a Polynomial

The polynomia...

Question

The polynomial $f (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$ when divided by $(x - 1)$ and $(x + 1)$ leaves the remainders $5$ and $9$ respectively. Find the values of $a$ and $b$ .

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Solution

$f (1) = 5$
$f (- 1) = 9$ $\to$ Remainder theorem
$f (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$
$f (1) = 1 - 2 + 3 - a + b = 5$
$2 - a + b = 5$
$b - a = 3$ ........... $(1)$
$f (- 1) = 9$
$1 + 2 + 3 + a + 6 = 9$
$6 + a + b = 9$
$a + b = 3$ …….. $(2)$
From $(1)$ and $(2)$
$a = 0$ , $b = 3$ .

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The polynomial f(x)=x4−2x3+3x2−ax+b when divided by (x−1) and (x+1)leaves the remainders 5 and 9 respectively. Find the values of a and b.

f(1)=5f(−1)=9 → Remainder theoremf(x)=x4−2x3+3x2−ax+bf(1)=1−2+3−a+b=52−a+b=5b−a=3 ...........(1)f(−1)=91+2+3+a+6=96+a+b=9a+b=3 ……..(2)From (1) and (2)a=0, b=3.

The polynomial $f (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$ when divided by $(x - 1)$ and $(x + 1)$ leaves the remainders $5$ and $9$ respectively. Find the values of $a$ and $b$ .

$f (1) = 5$
$f (- 1) = 9$ $\to$ Remainder theorem
$f (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$
$f (1) = 1 - 2 + 3 - a + b = 5$
$2 - a + b = 5$
$b - a = 3$ ........... $(1)$
$f (- 1) = 9$
$1 + 2 + 3 + a + 6 = 9$
$6 + a + b = 9$
$a + b = 3$ …….. $(2)$
From $(1)$ and $(2)$
$a = 0$ , $b = 3$ .