Example Question - difference of squares
Here are examples of questions we've helped users solve.
Polynomial Factorization Problem
<p>Để giải bài toán này, chúng ta sẽ tiến hành phân tích từng đa thức một:</p> <p>1/ \(x^4 + 6x^3 + 7x^2 - 6x + 1\)</p> <p>Đường tiếp cận thông thường là nhận thấy đây là dạng tổng của hình lập phương:</p> <p>\((x^2)^2 + 2 \cdot 3x(x^2) + (3x)^2 - 2 \cdot (x)(3x) + 1^2\)</p> <p>Điều này gợi nhớ đến công thức tổng của hai lập phương \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\), và ta có thể viết lại đa thức theo dạng:</p> <p>\((x^2 + 3x + 1)^2 - (2x \cdot 1)^2\)</p> <p>Điều này dẫn đến việc sử dụng công thức hiệu của hai bình phương \(a^2 - b^2 = (a+b)(a-b)\), từ đó ta có:</p> <p>\((x^2 + 3x + 1 + 2x)(x^2 + 3x + 1 - 2x)\)</p> <p>Đơn giản hóa các biểu thức:</p> <p>\((x^2 + 5x + 1)(x^2 + x + 1)\)</p> <p>2/ \( (x^2 + y^2 + z^2)(x + y + z)^2 - (xy + yz + zx)^2 \)</p> <p>Đây cũng là dạng của công thức hiệu của hai bình phương như trên:</p> <p>\((x + y + z)^2 - (xy + yz + zx)^2\)</p> <p>Ta có thể viết lại đa thức như sau:</p> <p>\((x + y + z + xy + yz + zx)(x + y + z - (xy + yz + zx))\)</p> <p>Đơn giản hóa:</p> <p>\((x + y + z + xy + yz + zx)(z - (xy + yz - zx))\)</p> <p>Biểu thức này không thể đơn giản hóa thêm mà không có thông tin cụ thể về \( x, y, z \), do đó đây là kết quả cuối cùng của phân tích.</p>
Factorization of a Difference of Squares with Higher Powers
La imagen muestra una expresión algebraica con la instrucción de factorizar. La expresión es: "8x^4 - y^4". Para factorizar la expresión, podemos reconocer que esto es una diferencia de cuadrados. La forma general para factorizar una diferencia de cuadrados es \( a^2 - b^2 = (a+b)(a-b) \). Sin embargo, puesto que tenemos cuartas potencias, aplicaremos la diferencia de cuadrados dos veces. Primero notamos que \( 8x^4 \) es \( (2x^2)^2 \) y que \( y^4 \) es \( y^2 \)^2. Por lo tanto, podemos escribir la expresión como una diferencia de cuadrados y aplicar la fórmula: \[ (2x^2)^2 - (y^2)^2 = (2x^2 + y^2)(2x^2 - y^2) \] Ahora, observamos que \( 2x^2 - y^2 \) también es una diferencia de cuadrados, que puede escribirse como \( ( \sqrt{2}x )^2 - y^2 \). Aplicamos nuevamente la fórmula de la diferencia de cuadrados: \[ 2x^2 - y^2 = ( \sqrt{2}x + y )( \sqrt{2}x - y ) \] Ahora substituimos esa factorización en nuestra expresión original: \[ (2x^2 + y^2)( \sqrt{2}x + y )( \sqrt{2}x - y ) \] Por lo tanto, la expresión \( 8x^4 - y^4 \) completamente factorizada es: \[ (2x^2 + y^2)( \sqrt{2}x + y )( \sqrt{2}x - y ) \]
Simplifying a Fraction Using Difference of Squares
The expression in the image appears to be a fraction that needs to be simplified: (1 - √2)/(√5 - √3) - (1 + √2)/(√5 + √3) To simplify this, we'll use the difference of squares technique. Here's the step-by-step process: 1. Multiply the numerator and denominator of each fraction by the conjugate of the respective denominator. For the first fraction, the conjugate of (√5 - √3) is (√5 + √3), and for the second fraction, the conjugate of (√5 + √3) is (√5 - √3). So, we have: (1 - √2)(√5 + √3) / [(√5 - √3)(√5 + √3)] - (1 + √2)(√5 - √3) / [(√5 + √3)(√5 - √3)] 2. Apply the difference of squares to the denominators: (√5)^2 - (√3)^2 = 5 - 3 = 2. 3. Expand the numerators: (1 - √2)(√5 + √3) = √5 + √3 - √10 - √6 (1 + √2)(√5 - √3) = √5 - √3 + √10 - √6 4. Now we have two fractions with the same denominator, which allows us to combine them: (√5 + √3 - √10 - √6) - (√5 - √3 + √10 - √6) / 2 5. Combine like terms in the numerator: The √5 and -√5 cancel each other out, the √10 and -√10 cancel each other out, and we're left with: (√3 - √3 - √6 + √6) / 2 6. Simplify the remaining terms: All terms in the numerator cancel out, leaving us with 0/2. 7. Therefore, the simplified form of the given expression is: 0 The entire simplified result is 0.
Solving Quadratic Equation by Factoring as Difference of Squares
The equation provided in the image is: \[100x^2 - 81 = 0\] This equation can be seen as a difference of squares, which takes the form \(a^2 - b^2 = (a+b)(a-b)\). Applying this to the given equation, we have: \[a^2 = (10x)^2\] \[b^2 = 81\] So the equation can be rewritten as: \[(10x)^2 - 9^2 = 0\] This further breaks down into: \[(10x + 9)(10x - 9) = 0\] Now, we solve each part of the equation separately to find the values of x: 1. \(10x + 9 = 0\) \[10x = -9\] \[x = -\frac{9}{10}\] 2. \(10x - 9 = 0\) \[10x = 9\] \[x = \frac{9}{10}\] Thus, the two solutions for \(x\) are \(-\frac{9}{10}\) and \(\frac{9}{10}\).
Factoring a Difference of Squares
The expression in the image is: \[ 100 - 121x^2 \] This is a difference of squares, which can be factored using the formula \( a^2 - b^2 = (a - b)(a + b) \). Here, \( a^2 = 100 \) and \( b^2 = 121x^2 \). Therefore, the square roots are \( a = 10 \) and \( b = 11x \). The factored form is: \[ (10 - 11x)(10 + 11x) \]
Algebraic Expression Factorization
The problem in the image is asking to factorize the given algebraic expression. Let's do this step-by-step. First, we have the expression: (x + 1) / (x - 2) + (x - 1) / (x^2 - 4) The second denominator, x^2 - 4, is a difference of squares and can be factored as: x^2 - 4 = (x + 2)(x - 2) Now let's rewrite the expression with this factored form: (x + 1) / (x - 2) + (x - 1) / [(x + 2)(x - 2)] To combine these fractions, they need to have a common denominator. The least common denominator is (x + 2)(x - 2). So, we multiply the numerator and denominator of the first fraction by (x + 2) to have the same denominator for both terms: [(x + 1)(x + 2)] / [(x - 2)(x + 2)] + (x - 1) / [(x + 2)(x - 2)] Expanding the numerator of the first term, we get: (x^2 + 3x + 2) / [(x - 2)(x + 2)] + (x - 1) / [(x + 2)(x - 2)] Now that we have a common denominator, we can combine the numerators: (x^2 + 3x + 2 + x - 1) / [(x + 2)(x - 2)] Combining like terms in the numerator: (x^2 + 4x + 1) / [(x + 2)(x - 2)] This is the final factorized form of the given expression. No further simplification is possible unless we further factor the numerator, which in this case, does not factor nicely over the integers or rationals.
Solving Rational Expressions with Factoring
To solve the equation \( \frac{3x^2 - 3}{x - 2x - 3} \), you first simplify the denominator: \( x - 2x - 3 = -x - 3 \) So the equation now is: \( \frac{3x^2 - 3}{-x - 3} \) Both the numerator and the denominator have a common factor which can be factored out. In the numerator, you can factor out a 3, and in the denominator, you can factor out a -1: \( \frac{3(x^2 - 1)}{-(x + 3)} \) Notice that \( x^2 - 1 \) is a difference of squares, which can be factored as \( (x + 1)(x - 1) \): \( \frac{3(x + 1)(x - 1)}{-(x + 3)} \) Depending on what you need to do next, the equation can be left in its factored form, or if you wish to simplify further, you can divide numerator and denominator by -1, which will change the signs: \( \frac{-3(x + 1)(x - 1)}{x + 3} \) Now the equation is fully simplified. If this is an expression that you are simplifying, that would be the final result. If you were solving for x, you would look if the denominator equals to zero because that would indicate where the expression is undefined. In this case, you cannot have \( x + 3 = 0 \), which simplifies to \( x \neq -3 \).
Factorization of Quadratic Expressions
To factorize the given expression in question 1, which is x^2 - 5x, you can factor out the greatest common factor (GCF), which in this case is x. Here's how: x^2 - 5x = x(x - 5) The expression is now factored as the product of x and (x - 5). For question 2, the expression to factorize is x^2 - 25. This is a difference of squares, which can be factored into the product of a sum and difference of the roots. Here's the factorization: x^2 - 25 = (x + 5)(x - 5)
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