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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

Px= 5x 5 + 3x...

Question

$P (x) = {5 x}^{5} + {3 x}^{3} - 7$ $g (x) = {2 x}^{2} + 1$ Find q(x) and r(x) by dividing p(x) by g(x)

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Solution

$P (x) = 5 x^{5} + 3 x^{3} - 7 g (x) = 2 x^{2} + 1$
(Image)
$∴ q (x) = \frac{5}{2} x^{3} + \frac{x}{2}$
$r (x) = \frac{- x}{2} - 7$

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Similar questions

Q. Who is correct?
Aakanksha:

q (x) = \frac{p (x) - r (x)}{g (x)}

q (x) = \frac{p (x)}{g (x)} - r (x)

q (x) = p (x) - \frac{r (x)}{g (x)}

q (x) = p (x) + \frac{r (x)}{g (x)}

p (x)

is the dividend,

g (x)

is the divisor,

q (x)

is the quotient, and

r (x)

is the remainder

Q. A polynomial

p (x)

g (x)

, the obtained quotient

q (x)

and the remainder

r (x)

are given in the table. Find

p (x)

Sl.	$p (x)$	$g (x)$	$q (x)$	$r (x)$
i	?	$x - 2$	$x^{2} - x + 1$	$4$
ii	?	$x + 3$	$2 x^{2} + x + 5$	$3 x + 1$
iii	?	$2 x + 1$	$x^{3} + 3 x^{2} - x + 1$	$0$
iv	?	$x - 1$	$x^{3} - x^{2} - x - 1$	$2 x - 4$
v	?	$x^{2} + 2 x + 1$	$x^{4} - 2 x^{2} + 5 x - 7$	$4 x + 12$

Q. Find the quotient

q (x)

r (x)

of the following when

f (x)

g (x)

.

p (x) = x^{6} + x^{4} - x^{2} - 1

;

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

p (x) = x^{3} + 8 x^{2} - 7 x + 12

g (x) = x - 1

p (x)

g (x)

q (x)

r (x)

as quotient and remainder respectively. If

a

is the degree of

q (x)

b

is the degree of

r (x)

(a - b) = ?

.

Q. On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, than p(x) = q(x)⋅g(x) + r(x), where

(a) r(x) = 0 always
(b) deg r (x) <deg g(x) always
(c) either r(x) = 0 or deg r(x) <deg g(x)
(d) r(x) = g(x)

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