Q. \(25x^{2}-16\).

Q: What is the factorization of 25x^2-16?

Recognize a difference of squares: 25x^2-16 = (5x)^2-4^2, so it factors as (5x-4)(5x+4).

Q: How do you solve 25x^2 - 16 = 0?.

Factor and set each factor to zero: (5x-4)(5x+4)=0 , so x=\frac{4}{5} or x=-\frac{4}{5} ..

Q: Is 25x^2 - 16 irreducible over the integers?

No. It factors over the integers as (5x-4)(5x+4), so it is reducible.

Q: What are the x-intercepts and vertex of y=25x^2-16 ?

X-intercepts at x=\frac{4}{5} and x=-\frac{4}{5}. Vertex at (0,-16) since the parabola y=25x^2-16 opens upward with axis x=0\ldots .

Q: How would you complete the square for 25x^2 - 16?

Factor 25: 25(x^2-16/25)=25[(x)^2-(4/5)^2]=25[(x-0)^2-(4/5)^2]. So completed-square form is 25[(x-0)^2-(4/5)^2]..

Q: What is the derivative and the integral of 25x^2-16?

What is the derivative and the integral of 25x^2-16?

Q: How does 25x^2-16 factor over the complex numbers?

The same as over the reals: 25x^2-16=(5x-4)(5x+4). Both linear factors have real (hence complex) roots x=\pm\frac{4}{5}..

Q: If coefficients changed, when is a quadratic a difference of squares?

Quadratic ax^2-b is a difference of squares when a and b are perfect squares (or can be written as squares), giving ax^2-b=\left(\sqrt{a}x-\sqrt{b}\right)\left(\sqrt{a}x+\sqrt{b}\right).

Answer

\( 25x^{2} – 16 \) is a difference of squares: \( (5x)^{2} – 4^{2} = (5x – 4)(5x + 4) \).

Final result: \( (5x – 4)(5x + 4) \)

Detailed Explanation

Problem

Simplify or factor the expression: \(25x^2 – 16\)

Solution — step-by-step

Recognize the form.The expression is a difference of two perfect squares because \(25x^2 = (5x)^2\) and \(16 = 4^2\). So rewrite the expression as\((5x)^2 – 4^2\)
Recall the difference-of-squares formula.For any expressions \(a\) and \(b\), the identity is\[a^2 – b^2 = (a – b)(a + b)\]
Apply the formula with \(a = 5x\) and \(b = 4\).Substitute into the identity to get\[(5x)^2 – 4^2 = (5x – 4)(5x + 4)\]
Verify by expanding the factors (optional but confirming).
1. Multiply the factors: \((5x – 4)(5x + 4)\).
2. Use distributive multiplication:\[(5x – 4)(5x + 4) = 5x\cdot 5x + 5x\cdot 4 – 4\cdot 5x – 4\cdot 4\]
3. Simplify each term:\[= 25x^2 + 20x – 20x – 16\]
4. Combine like terms \(20x – 20x = 0\):\[= 25x^2 – 16\]
This matches the original expression, confirming the factorization is correct.
Final answer.The factorization of \(25x^2 – 16\) is\[(5x – 4)(5x + 4)\]

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Algebra FAQs

What is the factorization of \(25x^2-16\)?

Recognize a difference of squares: \(25x^2-16 = (5x)^2-4^2\), so it factors as \((5x-4)(5x+4)\).

How do you solve \(25x^2 - 16 = 0\)?.

Factor and set each factor to zero: \( (5x-4)(5x+4)=0 \), so \( x=\frac{4}{5} \) or \( x=-\frac{4}{5} \)..

Is \(25x^2 - 16\) irreducible over the integers?

No. It factors over the integers as \((5x-4)(5x+4)\), so it is reducible.

What are the \(x\)-intercepts and vertex of \(y=25x^2-16\) ?

X-intercepts at \(x=\frac{4}{5}\) and \(x=-\frac{4}{5}\). Vertex at \((0,-16)\) since the parabola \(y=25x^2-16\) opens upward with axis \(x=0\ldots\) .

How would you complete the square for \(25x^2 - 16\)?

Factor 25: \(25(x^2-16/25)=25[(x)^2-(4/5)^2]=25[(x-0)^2-(4/5)^2]\). So completed-square form is \(25[(x-0)^2-(4/5)^2]\)..

What is the derivative and the integral of \(25x^2-16\)?

How does \(25x^2-16\) factor over the complex numbers?

The same as over the reals: \(25x^2-16=(5x-4)(5x+4)\). Both linear factors have real (hence complex) roots \(x=\pm\frac{4}{5}\)..

If coefficients changed, when is a quadratic a difference of squares?

Quadratic \(ax^2-b\) is a difference of squares when \(a\) and \(b\) are perfect squares (or can be written as squares), giving \(ax^2-b=\left(\sqrt{a}x-\sqrt{b}\right)\left(\sqrt{a}x+\sqrt{b}\right)\).

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