Pregunta
Exorcifo 1.14 You may have to rearrange and group the terms. Check some of your answers by multiplying the factors. Try to find more than one way of grouping for some of the questions. Factorise. \( \begin{array}{ll}\text { a) } a^{2}-b^{2}+a-b & \text { b) } p^{2}+p-q^{2}-q \\ \text { c) } 4 x^{2}+2 x-9 y^{2}+3 y & \text { d) } 5 a^{2} b^{2}-4 b^{4}+25 a^{4} b^{4}-2 b^{2} \\ \text { e) } 3 a^{2}-2 b-3 b^{2}+2 a & \text { f) } x^{3}-4 x y^{2}+x y-2 y^{2}\end{array} \)
Ask by Dunn Mcguire. in South Africa
Jan 23,2025
Solución de tutoría real
Respuesta verificada por el tutor
Responder
Here are the factorizations for each expression:
a) \( a^{2} - b^{2} + a - b = (a + b + 1)(a - b) \)
b) \( p^{2} + p - q^{2} - q = (p + q + 1)(p - q) \)
c) \( 4x^{2} + 2x - 9y^{2} + 3y = (2x + 3y)(2x - 3y + 1) \)
d) \( 5a^{2}b^{2} - 4b^{4} + 25a^{4}b^{4} - 2b^{2} = b^{2}(5a^{2}b^{2} + 2b^{2} + 1)(5a^{2} - 2) \)
e) \( 3a^{2} - 2b - 3b^{2} + 2a = (3a + 3b + 2)(a - b) \)
f) \( x^{3} - 4xy^{2} + xy - 2y^{2} = (x^{2} + 2xy + y)(x - 2y) \)
These factorizations can be verified by multiplying the factors back together to ensure they match the original expressions.
Solución
Factor the expression by following steps:
- step0: Factor:
\(a^{2}-b^{2}+a-b\)
- step1: Rewrite the expression:
\(a^{2}-ab+ba-b^{2}+a-b\)
- step2: Factor the expression:
\(a\left(a-b\right)+b\left(a-b\right)+a-b\)
- step3: Factor the expression:
\(\left(a+b+1\right)\left(a-b\right)\)
Factor the expression \( 4 x^{2}+2 x-9 y^{2}+3 y \).
Factor the expression by following steps:
- step0: Factor:
\(4x^{2}+2x-9y^{2}+3y\)
- step1: Rewrite the expression:
\(4x^{2}-6xy+2x+6yx-9y^{2}+3y\)
- step2: Factor the expression:
\(2x\left(2x-3y+1\right)+3y\left(2x-3y+1\right)\)
- step3: Factor the expression:
\(\left(2x+3y\right)\left(2x-3y+1\right)\)
Factor the expression \( 3 a^{2}-2 b-3 b^{2}+2 a \).
Factor the expression by following steps:
- step0: Factor:
\(3a^{2}-2b-3b^{2}+2a\)
- step1: Rewrite the expression:
\(3a^{2}-3ab+3ba-3b^{2}+2a-2b\)
- step2: Factor the expression:
\(3a\left(a-b\right)+3b\left(a-b\right)+2\left(a-b\right)\)
- step3: Factor the expression:
\(\left(3a+3b+2\right)\left(a-b\right)\)
Factor the expression \( x^{3}-4 x y^{2}+x y-2 y^{2 \).
Factor the expression by following steps:
- step0: Factor:
\(x^{3}-4xy^{2}+xy-2y^{2}\)
- step1: Rewrite the expression:
\(x^{3}-2x^{2}y+2x^{2}y-4xy^{2}+yx-2y^{2}\)
- step2: Factor the expression:
\(x^{2}\left(x-2y\right)+2xy\left(x-2y\right)+y\left(x-2y\right)\)
- step3: Factor the expression:
\(\left(x^{2}+2xy+y\right)\left(x-2y\right)\)
Factor the expression \( p^{2}+p-q^{2}-q \).
Factor the expression by following steps:
- step0: Factor:
\(p^{2}+p-q^{2}-q\)
- step1: Rewrite the expression:
\(p^{2}-pq+qp-q^{2}+p-q\)
- step2: Factor the expression:
\(p\left(p-q\right)+q\left(p-q\right)+p-q\)
- step3: Factor the expression:
\(\left(p+q+1\right)\left(p-q\right)\)
Factor the expression \( 5 a^{2} b^{2}-4 b^{4}+25 a^{4} b^{4}-2 b^{2 \).
Factor the expression by following steps:
- step0: Factor:
\(5a^{2}b^{2}-4b^{4}+25a^{4}b^{4}-2b^{2}\)
- step1: Evaluate:
\(25b^{4}a^{4}-4b^{4}+5b^{2}a^{2}-2b^{2}\)
- step2: Factor the expression:
\(b^{2}\left(25a^{4}b^{2}-4b^{2}+5a^{2}-2\right)\)
- step3: Factor the expression:
\(b^{2}\left(5a^{2}b^{2}+2b^{2}+1\right)\left(5a^{2}-2\right)\)
Here are the factorizations for each of the expressions provided:
### a) \( a^{2}-b^{2}+a-b \)
To factor this expression, we can rearrange and group the terms:
\[
a^{2} + a - b^{2} - b = (a + b + 1)(a - b)
\]
**Factorization:** \((a + b + 1)(a - b)\)
### b) \( p^{2}+p-q^{2}-q \)
Rearranging and grouping gives us:
\[
p^{2} + p - q^{2} - q = (p + q + 1)(p - q)
\]
**Factorization:** \((p + q + 1)(p - q)\)
### c) \( 4x^{2}+2x-9y^{2}+3y \)
Rearranging and grouping the terms:
\[
4x^{2} + 2x - 9y^{2} + 3y = (2x + 3y)(2x - 3y + 1)
\]
**Factorization:** \((2x + 3y)(2x - 3y + 1)\)
### d) \( 5a^{2}b^{2}-4b^{4}+25a^{4}b^{4}-2b^{2} \)
This expression can be factored as:
\[
5a^{2}b^{2} - 4b^{4} + 25a^{4}b^{4} - 2b^{2} = b^{2}(5a^{2}b^{2} + 2b^{2} + 1)(5a^{2} - 2)
\]
**Factorization:** \(b^{2}(5a^{2}b^{2} + 2b^{2} + 1)(5a^{2} - 2)\)
### e) \( 3a^{2}-2b-3b^{2}+2a \)
Rearranging and grouping gives:
\[
3a^{2} + 2a - 3b^{2} - 2b = (3a + 3b + 2)(a - b)
\]
**Factorization:** \((3a + 3b + 2)(a - b)\)
### f) \( x^{3}-4xy^{2}+xy-2y^{2} \)
Rearranging and grouping gives:
\[
x^{3} + xy - 4xy^{2} - 2y^{2} = (x^{2} + 2xy + y)(x - 2y)
\]
**Factorization:** \((x^{2} + 2xy + y)(x - 2y)\)
These factorizations can be verified by multiplying the factors back together to ensure they yield the original expressions.
Revisado y aprobado por el equipo de tutoría de UpStudy
Como
error msg
Explicar
Simplifique esta solución
Bonus Knowledge
Let’s dive into the fun world of factorization! For part (a): \(a^2 - b^2 + a - b\) can be rearranged to \((a^2 + a) - (b^2 + b)\) which factors to \(a(a + 1) - b(b + 1)\). It can also be expressed as \((a - b)(a + b + 1)\) through grouping. For part (b): \(p^2 + p - q^2 - q\) can be nicely grouped as \((p^2 - q^2) + (p - q)\). By factoring, \(p^2 - q^2\) gives \((p - q)(p + q)\) and thus overall it's \((p - q)(p + q + 1)\). Keep those factorization skills sharp!
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde
Hazte Premium