When x = 2 or x = 3, the value of polynomial q = x ^ 4 + ax ^ 3 + 32x ^ 2 + BX + 66 is 0. Find the quotient and remainder of Q / x ^ 2-5x + 6

When x = 2 or x = 3, the value of polynomial q = x ^ 4 + ax ^ 3 + 32x ^ 2 + BX + 66 is 0. Find the quotient and remainder of Q / x ^ 2-5x + 6

By substituting x = 2 and x = 3 into the polynomial Q, we get the following results:
2^4+a*2^3+32*2^2+b*2+66=0
3^4+a*3^3+32*3^2+b*3+66=0
The solution is: a = - 8; b = - 73
∴Q=x^4-8x^3+32x^2-73x+66
Q=x^4-8x^3+32x^2-73x+66
=x^4-5x^3+6x^2-3x^3+15x^2-18x+11x^2-55x+66
=x^2(x^2-5x+6)-3x(x^2-5x^2+6)+11(x^2-5x+6)
=(x^2-3x+11)(x^2-5x+6)
The quotient = x ^ 2-3x + 11
Remainder = 0

When x = 2 or x = 3, the value of polynomial q = x ^ 4 + ax ^ 3 + 32x ^ 2 + BX + 66 is 0. Try to find the product of polynomial Q and integral (3x + 1)

According to the meaning of the title:
Q=(X-2)(X-3)(X^2+mX+n)
The results are as follows
Q=X^4+(m-5)X^3+(n+6-5m)X^2+(6m-5n)X+6n
Set Q as the control question
a=m-5
n+6-5m=32
b=6m-5n
6n=66
The solution is as follows
m=-3
n=11
a=-8
b=-73
Therefore, the product of polynomial Q and integral (3x + 1) is:
(X-2)(X-3)(X^2-3X+11)(3X+1)

On Sunday, Xiaoming's mother went to the street to play, and saw a billboard at the door of a shop saying "all clothes in our shop are sold at a 20% discount". Xiaoming's mother took a fancy to one of the clothes. After some bargaining, the owner agreed to give another 10% discount. As a result, Xiaoming's mother spent 180 yuan to buy it. How much was the original price of this dress?

Let x, 80% X (1-10%) = 180 x = 250

In rectangle ABCD, ab = 2, BC = 4, if the rectangle is rotated around the edge AB, the volume of the enclosed geometry () is the same
The volume of the formed geometry is ()

PI times the square of four,

When the zeroth power of (x + 3) is 1, the value range of X is ()

Because there is no zero power of 0, x + 3 ≠ 0 and then x ≠ - 3
For any real number except 0, their zeroth power is 1
So the scope of this problem x is [x ≠ - 3]

As shown in the figure, a, B, C and D are the four vertices of the rectangle, ab = 16cm and ad = 6cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until it reaches B, and point Q moves to D at the speed of 2 & nbsp; cm / S. (1) how many seconds does P and Q start from the start? The area of quadrilateral pbcq is 33cm2; (2) from the start of P and Q to a few seconds? The distance between point P and point q is 10 cm

(1) Let P and Q have an area of 33cm2 from the start to x seconds, then Pb = (16-3x) cm, QC = 2xcm. According to the trapezoidal area formula, we get 12 (16-3x + 2x) × 6 = 33, and the solution is x = 5. (2) let P and Q have a distance of 10cm from the start to x seconds, then QE ⊥ AB, perpendicular foot e, QE = ad = 6, PQ = 10, ∵ PA = 3T, CQ = be = 2T, ∵ PE = ab-ap-be = | 16-5 T |, from Pythagorean theorem, we get (16-5t) 2 + 62 = 102, the solution is T1 = 4.8, T2 = 1.6. Answer: (1) the area of quadrilateral pbcq is 33cm2 when P and Q start from the starting point to 5 seconds; (2) when P and Q start from the starting point to 1.6 seconds or 4.8 seconds, the distance between P and Q is 10cm

As shown in the figure, it is known that RT △ ABC ≌ RT △ ade, ∠ ABC = ∠ ade = 90 °, BC and de intersect at point F and connect CD and EB. Please find a pair of congruent triangles in the figure and prove them

It is proved that: ∵ RT △ ABC ≌ RT △ ade, ≌ AC = AE, ab = ad, ∵ BAC = ≌ DAE, ≌ CAD = ≌ BAE, ≌ ACD ≌ AEB (SAS)

A car drives from the east to the West. After a journey, it is 210 kilometers away from the west, and then 20% of the whole journey. At this time, the ratio of the distance traveled to the distance not traveled is 3:2. How many kilometers are there between the East and the west?

Let x-210 + 20% x = 35x, x-210 + 0.2x = 0.6x, & nbsp; 1.2x-210 = 0.6x, & nbsp; 1.2x-0.6x = 210, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 0.6x = 210, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 350. A: the East and West cities are 350 km apart

An extremely difficult practical problem,
1. To assemble bicycles, 3 workers assemble 10 frames in 2 hours, 4 workers assemble 21 wheels in 3 hours, and 244 workers now. In order to assemble the price and wheels into a complete vehicle, how to arrange these workers?
There is also the second question. If the speed of the car from place a to place B is 5 kilometers slower than the scheduled speed per hour, the arrival time is 1 / 8 later than the scheduled time. If the speed is 1 / 3 higher than the scheduled speed, the arrival time is 1 hour earlier than the scheduled time, how about the distance between a and B?

The number of people needed to make a frame and wheel (same time)
(3×2÷10):(4×3÷21)=21:20
21/2=10.5
10.5:20=21:40
244÷(21+40)=4
Frame: 21 × 4 = 84 people
Round: 4 × 40 = 160 people

On a square board with a circumference of 80 cm, saw the largest circle. What is the area of the circle in square cm?

The side length of the square is: 80 △ 4 = 20 (CM), the area of the circle is: 3.14 × (20 △ 2) 2, = 3.14 × 102, = 3.14 × 100, = 314 (square cm). Answer: the area of the circle is 314 square cm