Problem 30
Question
Use a pattern to factor. Check. Identify any prime polynomials. $$ 64 m^{2}-49 p^{2} $$
Step-by-Step Solution
Verified Answer
(8m - 7p)(8m + 7p). Both factors are prime.
1Step 1: Recognize the form
The given expression is 64m^{2} - 49p^{2}. Notice that both terms are perfect squares and the expression is a difference of squares.
2Step 2: Write as squares
Express 64m^{2} and 49p^{2} as squares: (8m)^{2} - (7p)^{2}.
3Step 3: Apply the difference of squares formula
Use the difference of squares formula, which is a^{2} - b^{2} = (a - b)(a + b). Substituting 8m for a and 7p for b, the expression becomes (8m - 7p)(8m + 7p).
4Step 4: Verify the factored form
Expand (8m - 7p)(8m + 7p) to check: (8m)(8m) + (8m)(7p) - (7p)(8m) - (7p)(7p) = 64m^{2} - 49p^{2}, which is correct.
5Step 5: Check for prime polynomials
A prime polynomial cannot be factored further. Since (8m - 7p) and (8m + 7p) are binomials that do not share any common factors other than 1, they are considered prime polynomials.
Key Concepts
Difference of SquaresPrime PolynomialsPerfect Squares
Difference of Squares
A 'difference of squares' is a specific type of polynomial. It is expressed as the subtraction of two perfect squares. This form looks like:
a^{2} - b^{2}. For example, in the given exercise, 64m^{2} - 49p^{2}, we see two perfect squares subtracted.
To factor this, we use the difference of squares formula:
.The formula states that a² - b² can be factored into (a - b)(a + b).
By identifying 'a' and 'b' in our polynomial,
we see that a = 8m and b = 7p. Therefore, we apply the formula:
(8m)² - (7p)² = (8m - 7p)(8m + 7p). In the last step, we verify our answer by expanding, confirming we return to the original polynomial:
64m² - 49p². This confirms our factoring is correct.
a^{2} - b^{2}. For example, in the given exercise, 64m^{2} - 49p^{2}, we see two perfect squares subtracted.
To factor this, we use the difference of squares formula:
.The formula states that a² - b² can be factored into (a - b)(a + b).
By identifying 'a' and 'b' in our polynomial,
we see that a = 8m and b = 7p. Therefore, we apply the formula:
(8m)² - (7p)² = (8m - 7p)(8m + 7p). In the last step, we verify our answer by expanding, confirming we return to the original polynomial:
64m² - 49p². This confirms our factoring is correct.
Prime Polynomials
A 'prime polynomial' is like a prime number in arithmetic. It cannot be factored into simpler polynomials with integer coefficients.
After factoring 64m² - 49p² as (8m - 7p)(8m + 7p), we check each factor. In this case,
(8m - 7p) and (8m + 7p) cannot be factored further. This is because no common factor other than 1 exists.
Therefore, each factor is prime.
Identifying prime polynomials is essential, for it helps in simplifying expressions to their irreducible forms.
Remember, not all polynomials are prime. Always check if further factoring is possible.
After factoring 64m² - 49p² as (8m - 7p)(8m + 7p), we check each factor. In this case,
(8m - 7p) and (8m + 7p) cannot be factored further. This is because no common factor other than 1 exists.
Therefore, each factor is prime.
Identifying prime polynomials is essential, for it helps in simplifying expressions to their irreducible forms.
Remember, not all polynomials are prime. Always check if further factoring is possible.
Perfect Squares
A 'perfect square' is the square of an integer. Perfect squares are crucial for identifying patterns in polynomials. In the exercise provided, both terms in 64m² - 49p² are perfect squares.
For instance, 64m² is the square of 8m, and 49p² is the square of 7p. Recognition of perfect squares allows us to use specific factoring techniques, such as the difference of squares formula.
This insight simplifies factoring considerably. Perfect squares are also essential in solving quadratic equations and across various mathematical applications.
For instance, 64m² is the square of 8m, and 49p² is the square of 7p. Recognition of perfect squares allows us to use specific factoring techniques, such as the difference of squares formula.
This insight simplifies factoring considerably. Perfect squares are also essential in solving quadratic equations and across various mathematical applications.
Other exercises in this chapter
Problem 30
A student said that the solutions of \(3(x+8)=0\) are \(x=3\) and \(x=-8\). Explain what is wrong with this thinking.
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Factor completely. Identify any prime polynomials. $$ c^{2} d-15 c d+54 d $$
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(a) factor out the greatest common factor. Identify any prime polynomials. (b) check. $$ a^{5} b^{11}+a^{4} b^{10}+a^{3} b^{9}+a^{3} b^{5} $$
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Instead of using the zero product property, use the properties of equality to solve \(4(3 x+5)=0\).
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