Math Problem Statement

Given that p(x + 2)(x + 1) + q(x - 3)(x - 1) + r(x - 2)(x - 6) = 2x^2 + 7x - 13 for all values of x, find the values of p, q, and r.

Solution

We are given the equation:

p(x+2)(x+1)+q(x3)(x1)+r(x2)(x6)=2x2+7x13p(x + 2)(x + 1) + q(x - 3)(x - 1) + r(x - 2)(x - 6) = 2x^2 + 7x - 13

We need to find the values of pp, qq, and rr.

Step 1: Expand the terms on the left-hand side

Expand p(x+2)(x+1)p(x + 2)(x + 1):

p(x+2)(x+1)=p(x2+x+2x+2)=p(x2+3x+2)=px2+3px+2pp(x + 2)(x + 1) = p(x^2 + x + 2x + 2) = p(x^2 + 3x + 2) = p \cdot x^2 + 3p \cdot x + 2p

Expand q(x3)(x1)q(x - 3)(x - 1):

q(x3)(x1)=q(x2x3x+3)=q(x24x+3)=qx24qx+3qq(x - 3)(x - 1) = q(x^2 - x - 3x + 3) = q(x^2 - 4x + 3) = q \cdot x^2 - 4q \cdot x + 3q

Expand r(x2)(x6)r(x - 2)(x - 6):

r(x2)(x6)=r(x26x2x+12)=r(x28x+12)=rx28rx+12rr(x - 2)(x - 6) = r(x^2 - 6x - 2x + 12) = r(x^2 - 8x + 12) = r \cdot x^2 - 8r \cdot x + 12r

Step 2: Combine all expanded terms

Now, combine all terms on the left-hand side: px2+3px+2p+qx24qx+3q+rx28rx+12rp \cdot x^2 + 3p \cdot x + 2p + q \cdot x^2 - 4q \cdot x + 3q + r \cdot x^2 - 8r \cdot x + 12r Group the like terms (powers of xx): (p+q+r)x2+(3p4q8r)x+(2p+3q+12r)(p + q + r) \cdot x^2 + (3p - 4q - 8r) \cdot x + (2p + 3q + 12r)

Step 3: Set up equations by comparing coefficients

We now compare the coefficients of x2x^2, xx, and the constant terms with the right-hand side 2x2+7x132x^2 + 7x - 13.

  1. Coefficient of x2x^2: p+q+r=2p + q + r = 2
  2. Coefficient of xx: 3p4q8r=73p - 4q - 8r = 7
  3. Constant term: 2p+3q+12r=132p + 3q + 12r = -13

Step 4: Solve the system of equations

The system of equations is:

  1. p+q+r=2p + q + r = 2
  2. 3p4q8r=73p - 4q - 8r = 7
  3. 2p+3q+12r=132p + 3q + 12r = -13

Let me solve this system of equations.

From equation (1), solve for pp:

p=2qrp = 2 - q - r

Substitute this expression for pp into equations (2) and (3):

For equation (2): 3(2qr)4q8r=73(2 - q - r) - 4q - 8r = 7 63q3r4q8r=76 - 3q - 3r - 4q - 8r = 7 67q11r=76 - 7q - 11r = 7 7q11r=1(Equation 4)-7q - 11r = 1 \quad \text{(Equation 4)}

For equation (3): 2(2qr)+3q+12r=132(2 - q - r) + 3q + 12r = -13 42q2r+3q+12r=134 - 2q - 2r + 3q + 12r = -13 4+q+10r=134 + q + 10r = -13 q+10r=17(Equation 5)q + 10r = -17 \quad \text{(Equation 5)}

Step 5: Solve the two-variable system

We now have the system:

  1. 7q11r=1-7q - 11r = 1 (Equation 4)
  2. q+10r=17q + 10r = -17 (Equation 5)

Solve for qq in terms of rr from Equation (5): q=1710rq = -17 - 10r

Substitute this into Equation (4): 7(1710r)11r=1-7(-17 - 10r) - 11r = 1 119+70r11r=1119 + 70r - 11r = 1 119+59r=1119 + 59r = 1 59r=111959r = 1 - 119 59r=11859r = -118 r=2r = -2

Step 6: Solve for qq and pp

Substitute r=2r = -2 into Equation (5): q+10(2)=17q + 10(-2) = -17 q20=17q - 20 = -17 q=3q = 3

Now, substitute q=3q = 3 and r=2r = -2 into the expression for pp: p=23(2)=23+2=1p = 2 - 3 - (-2) = 2 - 3 + 2 = 1

Final Solution:

p=1,q=3,r=2p = 1, \quad q = 3, \quad r = -2

Would you like a more detailed breakdown or have any questions?


Here are some related questions for practice:

  1. What happens if we change the equation on the right-hand side?
  2. How would you approach solving a system with more unknowns?
  3. Can you find the discriminant of the quadratic expression for p=2p = 2?
  4. How does the solution change if the coefficients on the right-hand side are different?
  5. What method can we use to simplify solving this system?

Tip: When solving systems of equations, substitution is often effective when one variable is easy to isolate.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
System of Equations
Polynomial Expansion

Formulas

Expansion of (a + b)(c + d)
System of equations: comparing coefficients of powers of x

Theorems

Polynomial identity theorem
Solving systems of linear equations

Suitable Grade Level

Grades 9-12