Factor the following expression: 9y^4 - 7y^2 - 16
Understand the Problem
The question is asking to factor the expression (9y^4 - 7y^2 - 16). Factoring involves expressing the given polynomial as a product of simpler polynomials or factors. Here's how we can approach it.
Answer
$(y^2 + 1)(3y - 4)(3y + 4)$
Answer for screen readers
$(y^2 + 1)(3y - 4)(3y + 4)$
Steps to Solve
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Recognize the quadratic form Notice that the expression $9y^4 - 7y^2 - 16$ is in quadratic form. We can rewrite it as $9(y^2)^2 - 7(y^2) - 16$. Let $x = y^2$. This substitution gives us $9x^2 - 7x - 16$.
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Factor the quadratic We need to factor the quadratic expression $9x^2 - 7x - 16$. We are looking for two numbers that multiply to $9 \cdot -16 = -144$ and add up to $-7$. These numbers are $-16$ and $9$. Rewrite the middle term: $9x^2 - 16x + 9x - 16$ Factor by grouping: $x(9x - 16) + 1(9x - 16)$ $(x + 1)(9x - 16)$
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Substitute back $y^2$ for $x$ Replace $x$ with $y^2$ in the factored expression: $(y^2 + 1)(9y^2 - 16)$
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Factor the difference of squares Notice that $9y^2 - 16$ is a difference of squares, which can be factored as $(3y - 4)(3y + 4)$. So we have $(y^2 + 1)(3y - 4)(3y + 4)$
$(y^2 + 1)(3y - 4)(3y + 4)$
More Information
The factored form of the given expression is $(y^2 + 1)(3y - 4)(3y + 4)$. The term $(y^2+1)$ cannot be factored further using real numbers.
Tips
A common mistake is forgetting to substitute back $y^2$ for $x$ after factoring the quadratic. Another mistake is not recognizing and factoring the difference of squares.
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