Polynomials Test - 5

Length of the first side = (3x³ + 7x² – 8) m
Length of the second side = (5x³ – 9x² – 2) m
Perimeter = (12x³ – 8x² – 10) m

Now,

Perimeter = Length of the first side + Length of the second side + Length of the third side

(12x³ – 8x² – 10) = (3x³ + 7x² – 8) + (5x³ – 9x² – 2) + Length of the third side
(12x³ – 8x² – 10) = (8x³ – 2x² – 10) + Length of the third side

Length of the third side = (12x³ – 8x² – 10) - (8x³ – 2x² – 10)
= 12x³ – 8x² – 10 - 8x³ + 2x² + 10
= (4x³ - 6x²) m

Money spent in the first week = Rs. (10x³ + 5x² + 2)
For the second week:

Number of pencils she bought = (9x² + 10)
Cost of one pencil = Rs. (2x + 4)

Amount spent by her = Rs. (2x + 4)(9x² + 10) = Rs. (18x³ + 36x² + 20x + 40)
Money spent in two weeks altogether = Rs. {(10x³ + 5x² + 2) + (18x³ + 36x² + 20x + 40)}

Money spent in two weeks altogether = Rs. (28x³ + 41x² + 20x + 42)

Weight of one basket = (2x³ – 8x² + x – 4) kg = 2x²(x – 4) + 1(x – 4) = (x – 4)(2x² + 1)

Weight of another basket = (4x³ + 12x² + 2x + 6) kg = 2(2x³ + 6x² + x + 3) = 2(2x²(x + 3) + 1(x + 3)) = 2((x + 3)(2x² + 1))

Therefore, the required measure is the H.C.F. of both the measures, i.e. (2x² + 1) kg.

p(x) = x³ + 2x² - 4x + 7
g(x) = x + 2

By division algorithm,
p(x) = g(x)q(x) + r(x)
x³ + 2x² - 4x + 7 = (x + 2)(x² - a) + 15

Or, x³ + 2x² - 4x + 7 - 15 = (x + 2)(x² - a)
Or, x³ + 2x² - 4x - 8 = (x + 2)(x² - a)

Or, x²(x + 2) - 4(x + 2) = (x + 2)(x² - a)
Or, (x² - 4)(x + 2) = (x + 2)(x² - a)

Or, (x² - 4) = (x² - a)
Or, x² - x² = -a + 4
Or, a = 4

Let p(x) = x³ + 2x² - 5ax - 7.
Let R₁ be the remainder when p(x) is divided by x + 1.

R₁ = p(-1) = (-1)³ + 2(-1)² - 5(a)(-1) -7 = -1 + 2 + 5a - 7 = 5a - 6
Let q(x) = x³ + ax² - 12x + 6.

Let R₂ be the remainder when q(x) is divided by x - 2.
R₂ = q(2) = (2)³ + a(2)² -12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10

Now, 2R₁ + R₂ = 2 (5a - 6) + 4a - 10 = 6
Or 10a - 12 + 4a - 10 = 6

Or 14a - 22 = 6, 14a = 28
Or a = 28/14 = 2

2 and -2 are zeros of Qx⁵ – Px⁴ + Zx³ + Rx² + M.
Put x = 2

Q(2)⁵ – P(2)⁴ + Z(2)³ + R(2)² + M = 0
32Q – 16P + 8Z + 4R + M = 0 --- (1)

Now, put x = -2
Q(-2)⁵ – P(-2)⁴ + Z(-2)³ + R(-2)² + M = 0
-32Q – 16P - 8Z + 4R + M = 0 --- (2)

For (i)
Adding (1) and (2),
-32P + 8R + 2M = 0
8R + 2M = 32P

For (ii)
Subtracting (1) from (2),
64Q + 16Z = 0
4Q + Z = 0