Question
Which pattern can be used to factor the high-degree polynomial 16x^8-49x^2?
a. Neither the difference of squares nor the perfect square trinomial pattern can be used
b. (a-b)(a+b)=a^2-b^2
c. (a-b)^2=a^2-2ab+b^2
d. (a+b)^2=a^2+2ab+b^2
a. Neither the difference of squares nor the perfect square trinomial pattern can be used
b. (a-b)(a+b)=a^2-b^2
c. (a-b)^2=a^2-2ab+b^2
d. (a+b)^2=a^2+2ab+b^2
Answers
GPT-4o mini
To factor the given polynomial \( 16x^8 - 49x^2 \), we can first notice that it can be rewritten as:
\[
16x^8 - 49x^2 = (16x^8) - (49x^2)
\]
This expression is indeed a difference of squares, because it can be expressed as \( a^2 - b^2 \) where:
\[
a = (4x^4) \quad \text{and} \quad b = (7x)
\]
Thus, we recognize that:
\[
16x^8 - 49x^2 = (4x^4)^2 - (7x)^2
\]
Using the difference of squares factoring pattern \( a^2 - b^2 = (a - b)(a + b) \), we can factor this polynomial as follows:
\[
16x^8 - 49x^2 = (4x^4 - 7x)(4x^4 + 7x)
\]
Thus, the correct answer is option **b**: \((a-b)(a+b)=a^2-b^2\), as we used the difference of squares to factor the polynomial.
\[
16x^8 - 49x^2 = (16x^8) - (49x^2)
\]
This expression is indeed a difference of squares, because it can be expressed as \( a^2 - b^2 \) where:
\[
a = (4x^4) \quad \text{and} \quad b = (7x)
\]
Thus, we recognize that:
\[
16x^8 - 49x^2 = (4x^4)^2 - (7x)^2
\]
Using the difference of squares factoring pattern \( a^2 - b^2 = (a - b)(a + b) \), we can factor this polynomial as follows:
\[
16x^8 - 49x^2 = (4x^4 - 7x)(4x^4 + 7x)
\]
Thus, the correct answer is option **b**: \((a-b)(a+b)=a^2-b^2\), as we used the difference of squares to factor the polynomial.