Solution: Finding Values of a and b

Given Problem: If x3+ax2–bx+10x3+ax2–bx+10 is divisible by x2–3x+2x2–3x+2, we need to find the values of aa and bb.

Solution Steps:

Step 1: Determine the factors of the divisor

Given divisor: x2–3x+2x2–3x+2

We need to find two numbers that multiply to 22 and add up to −3−3.

The factors of 22 are 11 and 22.

So, the factors that add up to −3−3 are −2−2 and −1−1.

Hence, the divisor factors are (x–2)(x–2) and (x–1)(x–1).

So, the divisor can be written as (x–2)(x–1)(x–2)(x–1).

Step 2: Use Remainder Theorem

If f(x)=x3+ax2–bx+10f(x)=x3+ax2–bx+10 is divisible by (x–2)(x–1)(x–2)(x–1), then the remainder when f(x)f(x) is divided by x2–3x+2x2–3x+2 is zero.

According to Remainder Theorem, if f(x)f(x) is divided by x2–3x+2x2–3x+2, then the remainder is given by f(2)f(2) and f(1)f(1) respectively.

Step 3: Find the value of aa

Substitute x=2x=2 into f(x)f(x) and equate it to 00 to find the value of aa.

f(2)=23+a(2)2–b(2)+10f(2)=23+a(2)2–b(2)+10

0=8+4a–2b+100=8+4a–2b+10

18=4a–2b18=4a–2b

4a–2b=184a–2b=18

Step 4: Find the value of bb

Substitute x=1x=1 into f(x)f(x) and equate it to 00 to find the value of bb.

f(1)=13+a(1)2–b(1)+10f(1)=13+a(1)2–b(1)+10

0=1+a–b+100=1+a–b+10

11=a–b11=a–b

a–b=11a–b=11

Step 5: Solve the equations

Now we have two equations:

1. 4a–2b=184a–2b=18
2. a–b=11a–b=11

We can solve these equations simultaneously to find the values of aa and bb.

Step 6: Solve the equations

Equation 1: 4a–2b=184a–2b=18

Divide by 2: 2a–b=92a–b=9

Equation 2: a–b=11a–b=11

Step 7: Solve the system of equations

Adding equation 2 to equation 1: (2a–b)+(a–b)=9+11(2a–b)+(a–b)=9+11

3a=203a=20

a=203a=320

Substitute a=203a=320 into equation 2: 203–b=11320–b=11

b=203–11b=320–11

b=20–333b=320–33

b=−133b=3−13

Step 8: Final values of aa and bb

a=203a=320

b=−133b=3−13

So, the values of aa and bb are a=203a=320 and b=−133b=3−13 respectively.

Monomial and Binomial Examples with Degrees

Monomial Example (Degree: 82)

• Definition: A monomial is a mathematical expression consisting of a single term.
• Example: 5x825x82
  • Explanation:
    • The coefficient is 55.
    • The variable is xx.
    • The exponent is 8282.

Binomial Example (Degree: 99)

• Definition: A binomial is a polynomial with two terms.
• Example: 3x99+2x983x99+2x98
  • Explanation:
    • The first term: 3x993x99
      • Coefficient: 33
      • Variable: xx
      • Exponent: 9999
    • The second term: 2x982x98
      • Coefficient: 22
      • Variable: xx
      • Exponent: 9898

Additional Notes:

• Monomials have only one term, whereas binomials have two terms.
• The degree of a monomial is the sum of the exponents of its variables.
• The degree of a binomial is the highest degree of its terms.

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Heading 2: Step-by-Step Solution - Navigating the Equation Maze: To solve the problem, we follow a systematic approach:

Sub-heading 1: Expressing y in terms of x:

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• Substitute xy = 6 into the equation 3x + 2y = 12.
• Solve for y in terms of x.

Sub-heading 2: Substitution into the Expression:

• Once y is expressed in terms of x, substitute the value into the expression 9x² + 4y².
• This substitution simplifies the expression and facilitates further calculations.

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Perimeter Calculation for Rectangle with Given Area

Given Information:

• Area of the rectangle: 25x2−35x+1225x2−35x+12

Step 1: Determine the Dimensions

To calculate the perimeter of a rectangle, we need to know its length and width. We can find these dimensions using the area provided.

Step 2: Factorize the Area

Factorize the given quadratic expression 25x2−35x+1225x2−35x+12 to find its factors, which represent the possible lengths and widths of the rectangle.

Step 3: Use Factorization to Find Dimensions

Once the quadratic expression is factorized, identify the pairs of factors that, when multiplied, give the area of the rectangle. These pairs represent possible lengths and widths.

Step 4: Calculate Perimeter

With the length and width of the rectangle known, calculate the perimeter using the formula:

Perimeter=2×(Length+Width)Perimeter=2×(Length+Width)

Step 5: Finalize

Plug in the values of length and width into the perimeter formula to obtain the final result.

Let's proceed with these steps to find the perimeter.

Given:

• x2+y2+z2=83x2+y2+z2=83
• x+y+z=15x+y+z=15

To Find:

• x3+y3+z3−3xyzx3+y3+z3−3xyz

Approach:

1. Use the identity (x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x).
2. Given x+y+z=15x+y+z=15, find x3+y3+z3x3+y3+z3.
3. Also, find (x+y)(y+z)(z+x)(x+y)(y+z)(z+x).
4. Substitute the values in the expression x3+y3+z3−3xyzx3+y3+z3−3xyz.

Step-by-Step Solution:

1. Find x3+y3+z3x3+y3+z3:

   • Using the identity (x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x), where x+y+z=15x+y+z=15.
   • (15)3=x3+y3+z3+3(x+y)(y+z)(z+x)(15)3=x3+y3+z3+3(x+y)(y+z)(z+x)
   • 3375=x3+y3+z3+3(xy+yz+zx+3xyz)3375=x3+y3+z3+3(xy+yz+zx+3xyz) (Expanding (x+y+z)3(x+y+z)3)
   • x3+y3+z3=3375−3(xy+yz+zx)x3+y3+z3=3375−3(xy+yz+zx) (Subtracting 3xyz3xyz from both sides)

2. Find (x+y)(y+z)(z+x)(x+y)(y+z)(z+x):

   • Given x+y+z=15x+y+z=15, let's find xy+yz+zxxy+yz+zx.
   • Squaring x+y+z=15x+y+z=15:
     • (x+y+z)2=(15)2(x+y+z)2=(15)2
     • x2+y2+z2+2(xy+yz+zx)=225x2+y2+z2+2(xy+yz+zx)=225 (Expanding (x+y+z)2(x+y+z)2)
     • 83+2(xy+yz+zx)=22583+2(xy+yz+zx)=225 (Given x2+y2+z2=83x2+y2+z2=83)
     • xy+yz+zx=225−832=71xy+yz+zx=2225−83=71
   • Using (x+y)(y+z)(z+x)=(xy+yz+zx)+xyz(x+y)(y+z)(z+x)=(xy+yz+zx)+xyz:
     • (x+y)(y+z)(z+x)=71+xyz(x+y)(y+z)(z+x)=71+xyz

3. Substitute values into the expression:

   • x3+y3+z3−3xyz=3375−3(71)−3xyzx3+y3+z3−3xyz=3375−3(71)−3xyz
   • x3+y3+z3−3xyz=3375−213−3xyzx3+y3+z3−3xyz=3375−213−3xyz
   • x3+y3+z3−3xyz=3162−3xyzx3+y3+z3−3xyz=3162−3xyz

Final Answer:

• x3+y3+z3−3xyz=3162−3xyzx3+y3+z3−3xyz=3162−3xyz