class 9 polynomials extra questions

A.
1 -> x -1
2 -> $x^2 -x -1$
3 -> $1-3x^3$
4 -> $x^4 -3x^2+2 + 3x^3$
5 -> $x^5 - 3x^2 +1$
B.
Monomial : $3x^3$
Binomial : $x-1$
Trinomial : $x^2 -2x+1$
No apprpiate match : $x^4 -3x^2+2 + 3x^3$, $x^5 - 3x^2 +1$

P(x)=5x² -3x+7
P(0) = 5(0)² -3(0)+7=7
P(1) = 5(1)² -3(1)+7=9
P(5) = 5(5)² -3(5)+7=117
P(-1) = 5(-1)² -3(-1)+7=15
P(-2) = 5(-2)² -3(-2)+7=33

1. True, as g(1)=0
2. True, we can get this by split method
3. True
4. True
5. True , example x⁴ +1
6. True
7. True

(a) (x+6)(x+3)
(b) (3x-1)(x-1)(x+1)
(c)(x-1)(x-10)(x-12)
(d) (2x+1)(4x²-2x+1)

( c)

(d)

$\frac {x}{y} + \frac {y}{x}=-1$
$x^2 + y^2=-xy$
or
$x^2 +y^2 +xy=0$
Now
$x^3 – y^3= (x-y)(x^2 +y^2 +xy)=(x-y) \times 0=0$
Hence (c) is the correct answer

Now we know that
$p^3 + q^3 + r^3 -3pqr=(p+q+r)(p^2 + q^2 + r^2 -pq -qr -pr)$
as $p+q+r=0$
$p^3 + q^3 + r^3 -3pqr=0$ or $p^3 + q^3 + r^3 =3pqr$
Now
$\frac {p^2}{qr} + \frac {q^2}{pr} + \frac {r^2}{pq}= \frac {p^3 + q^3 + r^3}{pqr}= \frac {3pqr}{pqr}=3$
Hence (b) is the correct answer

$47^3 - 30^3 - 17^3=47^3 + (-30)^3 +(-17)^3$
Now we know that
$p^3 + q^3 + r^3 -3pqr=(p+q+r)(p^2 + q^2 + r^2 -pq -qr -pr)$
and if p+q+r=0,then
$p^3 + q^3 + r^3=3pqr$
Here 47+(-30)+ (-17)=0
Therefore
$47^3 - 30^3 - 17^3=47^3 + (-30)^3 +(-17)^3= 3 (47)(-30)(-17)=71910$

Now we know that
$p^3 + q^3 + r^3 -3pqr=(p+q+r)(p^2 + q^2 + r^2 -pq -qr -pr)$
and if p+q+r=0,then
$p^3 + q^3 + r^3=3pqr$
(i) Here $p^2 -q^2 + q^2 -r^2 + r^2 -p^2=0$
$ p -q + q -r + r -p=0$ Therefore
$\frac {(p^2 -q^2)^3 + (q^2 -r^2)^3 + (r^2 -p^2)^3}{(p -q)^3 + (q -r)^3 + (r -p)^3}$
$=\frac {3(p^2-q^2)(q^2 -r^2)(r^2 -p^2)}{3(p-q)(q-r)(r-p)}= (p+q)(q+r)(r+p)$
(ii) Here (x - y)+ (y – z)+ (z - x)=0
Therefore
$(x - y)^3+ (y – z)^3+ (z - x)^3=3(x-y)(y-z)(z-x)$

Let f(x) =x^3 -ax^2 -13x +b$
Now f(1)=0 and f(-3)=0
f(1) = 1-a -13+b=0 or b-a=12
f(-3)=9a-b-12=0
On solving both of these
a=3 and b=15