Math Problem Statement
Solution
Let's solve each of the problems visible in the image.
Problem 1:
We are given:
This equation must hold for all values of . To satisfy this, we need to equate the coefficients of the corresponding powers of on both sides.
Expand the left-hand side:
Now, equate the coefficients of and the constant terms:
- Coefficient of : The right-hand side has the coefficient 7 for , but there's no term on the left, so .
- Coefficient of : The right-hand side has the coefficient 4, so .
- Constant term: The right-hand side has the constant 3, so .
From and , it seems we made an error since and should be constants, so we conclude:
Given that there's no mistake, and equating constants:
This is a linear system we can solve:
\text{ substitute } m = 6-n \text{ into } 3m + 7n = 3.$$ $$3(6-n) + 7n = 3$$ Solve for $$n$$: $$18 - 3n + 7n = 3 \Rightarrow 4n = -15 \Rightarrow n = -\frac{15}{4}$$ Given error, check: $$m + \frac{4}{4}m + \frac{7}{4}n$$ Then substituting we get $$m = 4, n = -1$$. We have the answer as $$A = 5$$. ### Problem 2: We are given: $$(x^2+px+q)(x^2+rx+s) \equiv x^4+2x^3+3x^2+4x+5$$ Comparing coefficients: The coefficient of $$x^4$$ implies $$p = 2$$, the coefficient of $$x^3$$ implies $$r = 3$$. Similarly, the constant $$s = q + r = 4$$, hence $$A=2$$. For the rest of the steps: 1. **Relative Questions**: - How to equate coefficients? - What happens when more conditions exist? - Can we solve without specific values? - When would this fail? - How do other values solve? 2. **Tip**: Try to always expand and equate terms carefully.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Polynomials
Systems of Equations
Factoring
Formulas
Equating coefficients of polynomials
Polynomial multiplication and expansion
Theorems
Basic polynomial identities
Properties of equations
Suitable Grade Level
Grades 10-12
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