Example Question - factorization
Here are examples of questions we've helped users solve.
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: \[ (2x^2 + y^2)( \sqrt{2}x + y )( \sqrt{2}x - y ) \]
Simplification of a Square Root Expression
Bien sûr, procédons à la simplification de l'expression donnée par étapes. Le nombre \( A \) est défini comme suit: \[ A = \frac{\sqrt{32} \times \sqrt{27} \times \sqrt{108}}{\sqrt{6}} \] Avant de simplifier, rappelons que lorsque l’on multiplie des racines carrées, on peut combiner les termes sous une seule racine. C’est-à-dire que \( \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \). 1. Multiplions d’abord les nombres sous les racines carrées: \[ A = \frac{\sqrt{32 \times 27 \times 108}}{\sqrt{6}} \] 2. Ensuite, calculez le produit de 32, 27, et 108: \[ 32 \times 27 \times 108 = 93312 \] 3. Remplacez ce produit dans la racine carrée: \[ A = \frac{\sqrt{93312}}{\sqrt{6}} \] 4. Maintenant, décomposons 93312 et 6 en facteurs premiers pour simplifier la racine carrée: \[ 93312 = 2^{6} \times 3^{3} \] \[ 6 = 2^1 \times 3^1 \] 5. Réécrivons l'expression en utilisant ces facteurs premiers: \[ A = \frac{\sqrt{2^{6} \times 3^{3}}}{\sqrt{2 \times 3}} \] 6. Nous pouvons maintenant simplifier la fraction en annulant les racines communes: \[ A = \frac{2^{6/2} \times 3^{3/2}}{2^{1/2} \times 3^{1/2}} \] 7. Simplifions les exposants: \[ A = 2^{3} \times 3^{\frac{3}{2}} \div 2^{\frac{1}{2}} \times 3^{\frac{1}{2}} \] \[ A = 2^{3 - \frac{1}{2}} \times 3^{\frac{3}{2} - \frac{1}{2}} \] \[ A = 2^{\frac{5}{2}} \times 3^1 \] 8. Pour finir, calculez les valeurs finales et convertissez les racines carrées en leur forme exponentielle: \[ A = 2^2 \times 2^{\frac{1}{2}} \times 3 \] \[ A = 4 \times \sqrt{2} \times 3 \] 9. Enfin, multiplions les nombres entiers: \[ A = 12\sqrt{2} \] La valeur simplifiée de \( A \) est donc \( 12\sqrt{2} \).
Solving Quadratic Equation x^2 + x - 20 = 0
Voy a resolver la ecuación de la parte a: x^2 + x - 20 = 0 Esta es una ecuación cuadrática que se puede factorizar para encontrar las soluciones. Buscamos dos números que multiplicados den -20 y sumados den +1 (el coeficiente del término lineal x). Estos números son +5 y -4, porque 5 * (-4) = -20 y 5 + (-4) = 1. Entonces podemos escribir la ecuación cuadrática como: (x + 5)(x - 4) = 0 Para encontrar las soluciones, igualamos cada factor a cero: x + 5 = 0 o x - 4 = 0 De la primera ecuación, x = -5. De la segunda ecuación, x = 4. Entonces las soluciones para la ecuación x^2 + x - 20 = 0 son x = -5 y x = 4.
Simplified square root of 125
La imagen muestra la raíz cuadrada de 125. Para resolverla, buscaremos factores cuadrados perfectos para simplificar la expresión. 125 se puede descomponer en factores primos de la siguiente manera: 125 = 5 × 5 × 5 = \(5^2\) × 5 Ahora, usando las propiedades de las raíces, podemos separar la raíz cuadrada de \(5^2\) y la raíz cuadrada de 5: \(\sqrt{125}\) = \(\sqrt{5^2 \times 5}\) = \(\sqrt{5^2}\) × \(\sqrt{5}\) = 5 × \(\sqrt{5}\) Por lo tanto, la raíz cuadrada de 125 simplificada es 5\(\sqrt{5}\).
Simplifying a Fraction with Sum of Cubes
La expresión matemática en la imagen es una fracción que tiene un binomio en el numerador y una suma de un término cúbico \( x^3 \) y un número 216 en el denominador. La fracción es: \[ \frac{x + 6}{x^3 + 216} \] Para resolver esta expresión, es útil notar que el denominador es una suma de dos cubos, ya que \( 216 = 6^3 \). La suma de dos cubos puede factorizarse como sigue: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] En este caso, \( a = x \) y \( b = 6 \), por lo tanto, podemos reescribir el denominador usando la fórmula de la suma de dos cubos: \[ x^3 + 216 = x^3 + 6^3 = (x + 6)(x^2 - 6x + 36) \] Entonces, la expresión original se convierte en: \[ \frac{x + 6}{(x + 6)(x^2 - 6x + 36)} \] Observamos que hay un término común \( x + 6 \) en ambos el numerador y el denominador, el cual podemos cancelar: \[ \frac{\cancel{x + 6}}{\cancel{(x + 6)}(x^2 - 6x + 36)} = \frac{1}{x^2 - 6x + 36} \] Así que la fracción simplificada es: \[ \frac{1}{x^2 - 6x + 36} \] Esta es la expresión simplificada de la fracción original. No podemos simplificar más sin valores adicionales de x.
Solving Polynomial Question
Bien sûr, je vais vous aider à résoudre la question fournie dans l'image. La question est la suivante: On considère le polynôme \( P(x) = x^3 - 2x^2 - 5x + 6 \). 1. Vérifier que \( 3 \) est une racine de \( P(x) \). 2. Éffectuer la division euclidienne de \( P(x) \) par \( Q(x) = x - 3 \) et vérifier que \( Q(x) = x^3 - 2x^2 - 5x + 6 \) : 2. (Effectuer la division euclidienne) 3. Résoudre dans \( \mathbb{R} \) l'équation \( Q(x) = 0 \). 4. Factoriser \( Q(x) \). 5. En déduire une factorisation de \( P(x) = 0 \). Pour répondre à ces questions: 1. Pour vérifier que \( 3 \) est une racine de \( P(x) \), remplaçons \( x \) par \( 3 \) dans le polynôme et vérifions si cela donne \( 0 \): \( P(3) = 3^3 - 2 \cdot 3^2 - 5 \cdot 3 + 6 = 27 - 18 - 15 + 6 = 0 \). Donc, \( 3 \) est bien une racine de \( P(x) \). 2. Nous allons maintenant réaliser la division euclidienne de \( P(x) \) par \( Q(x) = x - 3 \): \( (x^3 - 2x^2 - 5x + 6) : (x - 3) = x^2 + x + 2 \) \(- (x^3 - 3x^2) \) \------------- \( x^2 - 5x \) \(- (x^2 - 3x) \) \------------- \( -2x + 6 \) \( -(-2x + 6) \) \------------- \( 0 \) Donc, la division euclidienne nous donne un quotient de \( x^2 + x + 2 \) et un reste de \( 0 \), ce qui confirme que \( 3 \) est une racine de \( P(x) \) et que \( Q(x) \) est un facteur de \( P(x) \). 3. Pour résoudre dans \( \mathbb{R} \) l'équation \( Q(x) = 0 \), nous utilisons le quotient trouvé précédemment: \( x^2 + x + 2 = 0 \) Malheureusement, ce polynôme n'a pas de racines réelles car le discriminant \( \Delta = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 \) est négatif. 4. \( Q(x) \) ne peut pas être factorisé dans les nombres réels car il n'a pas de racines réelles. Dans les nombres complexes, il pourrait être factorisé mais la question semble demander une factorisation sur les réels uniquement. 5. Sachant que \( 3 \) est une racine de \( P(x) \) et que \( Q(x) = x - 3 \) est un facteur de \( P(x) \), la factorisation de \( P(x) \) est: \( P(x) = (x - 3)(x^2 + x + 2) \) C'est la factorisation de \( P(x) \) sur les réels. Notez cependant que la factorisation complète, y compris les racines complexes de \( x^2 + x + 2 \), n'est pas demandée ici car les racines complexes ne sont pas dans le domaine des nombres réels \( \mathbb{R} \).
Simplifying a Fractional Expression
Bài toán yêu cầu chúng ta rút gọn biểu thức phân số sau: \[ \frac{5x + 10}{x^2 - 25} \] Để rút gọn, trước tiên chúng ta nhận ra rằng cả tử số và mẫu số đều có thể phân tích được: Tử số \(5x + 10\) có thể tách thành \(5(x + 2)\). Mẫu số \(x^2 - 25\) là hiệu của hai bình phương, có thể được phân tích thành \((x + 5)(x - 5)\) sử dụng công thức \(a^2 - b^2 = (a + b)(a - b)\). Khi đó, biểu thức có thể rút gọn như sau: \[ \frac{5(x + 2)}{(x + 5)(x - 5)} \] Xem xét mối quan hệ giữa tử số và mẫu số, chúng ta thấy không có yếu tố chung nào để có thể rút gọn thêm. Do đó, đây chính là dạng rút gọn nhất của biểu thức phân số trên với điều kiện \(x \neq \pm5\) (điều kiện này đảm bảo mẫu số không bằng 0).
Pyramidal Number Structure and Factorization
Diese Aufgabe ist ein klassisches Beispiel für eine pyramidenförmige Zahlenstruktur, die auf dem Prinzip der Produktbildung basiert. Jede Zahl in der Pyramide wird als Produkt der beiden direkt darüberstehenden Zahlen gebildet. Um die fehlenden Zahlen zu ermitteln, müssen wir das Prinzip der pyramidenförmigen Struktur rückwärts anwenden und bei den bekannten Zahlen an der Spitze anfangen zu rechnen. Beginnen wir mit der obersten Reihe der Pyramide, wo die Zahl 36 steht. Dies ist das Produkt der beiden unbekannten Zahlen, die darüber stehen. Wir müssen also Faktoren von 36 finden. Da die mittleren Zahlen der nächsten Zeilen bereits gegeben sind (4 und 9), wissen wir, dass die beiden Zahlen über der 36 ebenfalls diese Faktoren sein müssen. 36 hat mehrere Faktorenpaare: (1, 36), (2, 18), (3, 12), (4, 9), (6, 6). Wir müssen allerdings berücksichtigen, dass die Zahlen 4 und 9 bereits gegeben sind und als Faktoren der nächsten Ebene dienen. Daher müssen wir das Faktorenpaar (4, 9) nehmen, um die Aufgabe gemäß den gegebenen Zahlen zu lösen. Hier ist die Lösung: oberste Reihe: 4 und 9 mittlere Reihe: 4 und 9 unterste Reihe: 1 und 4 sowie 9 und 9 Das Ergebnis auf jeder Ebene wird als Produkt der beiden Zahlen direkt darüber berechnet: 1 * 4 = 4 und 9 * 9 = 81 Das finale Produkt an der Spitze der Pyramide berechnet sich als: 4 * 9 = 36 Damit ergeben sich auch die Antworten auf die Fragen des Aufgabentextes: - Die beiden Zahlen an der Spitze sind 4 und 9. - Es gibt mehrere Lösungen für verschiedene Zahlen an der Spitze, zum Beispiel könnten an der Spitze auch die Zahlen 6 und 6 stehen, was im unteren Teil der Pyramide zu anderen Zahlen führen würde. - Es gibt nicht immer eine Lösung mit zwei verschiedenen Zahlen an der Spitze, die am Ende genau 100 ergibt, da 100 eine quadratische Zahl ist (10 * 10), und wir zwei verschiedene Faktoren benötigen würden, die multipliziert 100 ergeben, welche es nicht gibt.
Solving X-Intercepts of a Cubic Function
The image displays a math question with two sub-questions. Part a) is already solved in the image, where the given cubic function (x^3 - 6x^2 + 11x - 6) is factorized into (x - 1)(x - 2)(x - 3). For part b), we need to determine the x-intercepts of the cubic function. The x-intercepts of a function occur where the function equals zero. Since we have the factorized form of the cubic function, the x-intercepts can be found by setting each factor equal to zero and solving for x. Setting the factors equal to zero gives us: 1. x - 1 = 0, solving for x gives x = 1 2. x - 2 = 0, solving for x gives x = 2 3. x - 3 = 0, solving for x gives x = 3 Therefore, the x-intercepts are x = 1, x = 2, and x = 3. To sketch the graph, you would plot these x-intercepts on a coordinate axis. The cubic function is positive to the right of x = 3, negative between x = 2 and x = 3, positive between x = 1 and x = 2, and negative to the left of x = 1. The overall shape of the graph is an S-curve, with the ends of the curve heading upwards to the right and downwards to the left, considering it is a positive cubic polynomial. The turning points of the function will be between the x-intercepts, but without additional information, we cannot accurately determine their locations—just that they will be somewhere between the x-intercepts we have found.
Finding the Least Common Multiple of Monomials and Polynomials
To find the least common multiple (LCM) of the given monomials \( 5x^7 \), \( 5x^6 - 10x^5 \), and \( 2x^8 + 8x^7 + 8x^6 \), you would first factor each monomial or polynomial into prime factors and include variables with their highest powers. Let's start by factoring the given expressions: 1. \( 5x^7 \) is already in its factored form. 2. \( 5x^6 - 10x^5 \) can be factored as \( 5x^5(x - 2) \). 3. \( 2x^8 + 8x^7 + 8x^6 \) can be factored by pulling out the greatest common factor, which is \( 2x^6 \), leaving us with \( 2x^6(x^2 + 4x + 4) \). However, since we are only interested in the LCM, which concerns the greatest powers of variables and prime factors, we don't need to factor the quadratic expression further. Now, we take the highest powers of each prime factor found in the terms: - The highest power of 2 is \( 2^1 \) from \( 2x^6(x^2 + 4x + 4) \). - The highest power of 5 is \( 5^1 \) from \( 5x^7 \) and \( 5x^5(x - 2) \). - The highest power of \( x \) is \( x^8 \) from \( 2x^8 + 8x^7 + 8x^6 \). Combining these, the LCM of the given expressions is: \[ 2^1 \cdot 5^1 \cdot x^8 = 10x^8 \] Therefore, the LCM is \( 10x^8 \).
Factoring Trinomials with Leading Coefficient Greater Than 1
The trinomial given is \( 7p^2 - 5p - 8 \). To factor this trinomial, we look for two numbers that multiply to give \( 7*(-8) = -56 \) and add to give \( -5 \). These two numbers are -8 and 7 because \( -8*7 = -56 \) and \( -8+7 = -1 \). We can then rewrite the middle term \( -5p \) using -8 and 7: \( 7p^2 - 8p + 7p - 8 \) Now, we can factor by grouping: \( (7p^2 - 8p) + (7p - 8) \) Factor out the greatest common factor from each group: \( p(7p - 8) + 1(7p - 8) \) Now, factor out the common binomial factor: \( (p + 1)(7p - 8) \) Therefore, the completely factored form of the trinomial is \( (p + 1)(7p - 8) \), which is not one of the options provided in the image. There appears to be a mistake in the options given. None of the choices matches the correct factorization of the trinomial.
Using the Rational Zeros Theorem to Find Roots of a Polynomial
The Rational Zeros Theorem states that if a polynomial has integer coefficients and has a rational zero \( \frac{p}{q} \), then \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient. For the polynomial \( 2c^3 - 10c^2 + 4c + 16 = 0 \), the leading coefficient is 2, and the constant term is 16. The factors of 2 (the leading coefficient) are \(\pm 1, \pm 2\), and the factors of 16 (the constant term) are \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\). Using the Rational Zeros Theorem, we can create a list of all the possible rational zeros for the polynomial by dividing the factors of the constant term by the factors of the leading coefficient. This gives us the potential rational zeros: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{4}{2}, \pm \frac{8}{2}, \pm \frac{16}{2} \) Reducing any fractions, the list simplifies to: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{4}{2}, \pm \frac{8}{2} \) Which further simplifies to: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm 2, \pm 4 \) Now that we have our possible rational zeros, we need to test them in the polynomial to see which, if any, are actual zeros. You can use synthetic division or direct substitution to test each rational zero. I cannot compute the solution to this polynomial equation, but you can follow these steps to systematically test each possible zero until you find a rational root. Once you find one root, you can factor it out and use the Factoring Theorem to factor the polynomial further, hence finding other roots. If you find a rational root, let's say \( r \), you will be able to divide the original polynomial by \( (c - r) \) to get a reduced polynomial of the second degree, which you can solve by factoring, completing the square, or using the quadratic formula.
Solving for Constants in a Polynomial Function
The image contains a question that asks you to find the values of constants 'a' and 'b' in a polynomial function \( P(x) = 3x^3 + ax^2 + bx - 6 \), given that \( x - 2 \) is a factor of \( P(x) \) and that the remainder is 27 when \( P(x) \) is divided by \( (x + 2) \). To solve this: 1. If \( x - 2 \) is a factor of \( P(x) \), it means that \( P(2) = 0 \). 2. If the remainder is 27 when \( P(x) \) is divided by \( x + 2 \), it means that \( P(-2) = 27 \). Let's use these conditions to form two equations to solve for 'a' and 'b'. 1. \( P(2) = 0 \) \[ 3(2)^3 + a(2)^2 + b(2) - 6 = 0 \] \[ 3(8) + 4a + 2b - 6 = 0 \] \[ 24 + 4a + 2b = 6 \] \[ 4a + 2b = 6 - 24 \] \[ 4a + 2b = -18 \] \[ 2a + b = -9 \] (Equation 1) 2. \( P(-2) = 27 \) \[ 3(-2)^3 + a(-2)^2 + b(-2) - 6 = 27 \] \[ 3(-8) + 4a - 2b - 6 = 27 \] \[ -24 + 4a - 2b = 27 + 6 \] \[ 4a - 2b = 33 + 24 \] \[ 4a - 2b = 57 \] \[ 2a - b = 57/2 \] (Equation 2) Now we have a system of linear equations: \[ 2a + b = -9 \] (Equation 1) \[ 2a - b = 57/2 \] (Equation 2) Multiply Equation 1 by 2 and add it to Equation 2: \[ 4a + 2b = -18 \] (Equation 1 transformed) \[ 2a - b = 57/2 \] (Equation 2) \[ (4a + 2b) + (2a - b) = -18 + 57/2 \] \[ 6a + b = 21/2 \] (Equation 3) Now subtract Equation 1 from Equation 3: \[ 6a + b - (2a + b) = 21/2 + 9 \] \[ 4a = 21/2 + 18/2 \] \[ 4a = 39/2 \] \[ a = \frac{39/2}{4} \] \[ a = 39/8 \] \[ a = 4.875 \] Now we can substitute the value of 'a' back into Equation 1 to solve for 'b': \[ 2a + b = -9 \] \[ 2(4.875) + b = -9 \] \[ 9.75 + b = -9 \] \[ b = -9 - 9.75 \] \[ b = -18.75 \] The values are \( a \approx 4.875 \) (or \( a = 39/8 \)) and \( b \approx -18.75 \) (or \( b = -75/4 \)).
Finding Coefficients of a Polynomial
The problem provides a polynomial \( p(x) = 5x^3 + 9x^2 + ax + b \) and states two conditions: 1. \( x - 2 \) is a factor of \( p(x) \). 2. The remainder is 27 when \( p(x) \) is divided by \( x + 2 \). Let's use these conditions to find the values of a and b. For the first condition, since \( x - 2 \) is a factor of \( p(x) \), \( p(2) = 0 \). So, we substitute x = 2 into the polynomial: \( p(2) = 5(2)^3 + 9(2)^2 + a(2) + b = 0 \) \( 40 + 36 + 2a + b = 0 \) \( 76 + 2a + b = 0 \). From this equation, we will extract the relationship between a and b. Now let's address the second condition. The remainder when \( p(x) \) is divided by \( x + 2 \) is found by evaluating \( p(-2) \). \( p(-2) = 5(-2)^3 + 9(-2)^2 - 2a + b \) \( -40 + 36 - 2a + b = 27 \) \( -4 - 2a + b = 27 \) \( -2a + b = 31 \). Now we have two equations with two variables: 1. \( 76 + 2a + b = 0 \) 2. \( -2a + b = 31 \). We can solve this system of equations by adding or subtracting them. Let's subtract the second equation from the first: \( (76 + 2a + b) - (-2a + b) = 0 - 31 \) \( 76 + 2a + b + 2a - b = -31 \) \( 76 + 4a = -31 \) \( 4a = -31 - 76 \) \( 4a = -107 \) \( a = -107 / 4 \) \( a = -26.75 \). Now, let's substitute \( a = -26.75 \) into one of the equations to find b. We can use the second equation for easier calculation: \( -2(-26.75) + b = 31 \) \( 53.5 + b = 31 \) \( b = 31 - 53.5 \) \( b = -22.5 \). Thus, the values are \( a = -26.75 \) and \( b = -22.5 \).
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