Login
Premium
Home Question Bank Math Algebra
Question
upstudy study bank question image url

Expand and simplify: \( \begin{array}{lll}\text { (1) } 2 x(3 x-4 y)-\left(7 x^{2}-2 x y\right) & \text { (2) } \quad(5 y+1)^{2}-(3 y+4)(2-3 y) \\ \text { (3) }(2 x+y)^{2}-(3 x-2 y)^{2}+(x-4 y)(x+4 y) \\ \text { (4) } x^{6}+\left(x^{3}-3 y\right)\left(x^{3}+3 y\right) & \text { (5) } \quad(3 a+b)(3 a-b)(2 a+5 b)\end{array} \)

Ask by Clark Logan. in South Africa
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

1. \(-x^{2} - 6xy\) 2. \(34y^{2} + 16y - 7\) 3. \(-4x^{2} + 16xy - 19y^{2}\) 4. \(2x^{6} - 9y^{2}\) 5. \(18a^{3} + 45a^{2}b - 2ab^{2} - 5b^{3}\)

Solution

Let's expand and simplify each expression step by step. --- **1. \( 2x(3x - 4y) - (7x^{2} - 2xy) \)** **Expansion:** \[ 2x(3x) + 2x(-4y) - 7x^{2} + 2xy = 6x^{2} - 8xy - 7x^{2} + 2xy \] **Combine like terms:** \[ (6x^{2} - 7x^{2}) + (-8xy + 2xy) = -x^{2} - 6xy \] **Simplified Form:** \[ - x^{2} - 6xy \] --- **2. \( (5y + 1)^{2} - (3y + 4)(2 - 3y) \)** **Expansion:** \[ (5y)^{2} + 2 \cdot 5y \cdot 1 + 1^{2} - [3y \cdot 2 + 3y \cdot (-3y) + 4 \cdot 2 + 4 \cdot (-3y)] \] \[ = 25y^{2} + 10y + 1 - (6y - 9y^{2} + 8 - 12y) \] \[ = 25y^{2} + 10y + 1 - 6y + 9y^{2} - 8 + 12y \] **Combine like terms:** \[ (25y^{2} + 9y^{2}) + (10y - 6y + 12y) + (1 - 8) = 34y^{2} + 16y - 7 \] **Simplified Form:** \[ 34y^{2} + 16y - 7 \] --- **3. \( (2x + y)^{2} - (3x - 2y)^{2} + (x - 4y)(x + 4y) \)** **Expansion:** \[ (2x)^{2} + 2 \cdot 2x \cdot y + y^{2} - [(3x)^{2} + 2 \cdot 3x \cdot (-2y) + (-2y)^{2}] + (x^{2} - (4y)^{2}) \] \[ = 4x^{2} + 4xy + y^{2} - (9x^{2} - 12xy + 4y^{2}) + x^{2} - 16y^{2} \] \[ = 4x^{2} + 4xy + y^{2} - 9x^{2} + 12xy - 4y^{2} + x^{2} - 16y^{2} \] **Combine like terms:** \[ (4x^{2} - 9x^{2} + x^{2}) + (4xy + 12xy) + (y^{2} - 4y^{2} - 16y^{2}) = -4x^{2} + 16xy - 19y^{2} \] **Simplified Form:** \[ -4x^{2} + 16xy - 19y^{2} \] --- **4. \( x^{6} + (x^{3} - 3y)(x^{3} + 3y) \)** **Expansion using the difference of squares:** \[ (x^{3} - 3y)(x^{3} + 3y) = x^{6} - (3y)^{2} = x^{6} - 9y^{2} \] **Combine with the existing \( x^{6} \):** \[ x^{6} + x^{6} - 9y^{2} = 2x^{6} - 9y^{2} \] **Simplified Form:** \[ 2x^{6} - 9y^{2} \] --- **5. \( (3a + b)(3a - b)(2a + 5b) \)** **First, expand \( (3a + b)(3a - b) \) using the difference of squares:** \[ (3a + b)(3a - b) = 9a^{2} - b^{2} \] **Now, multiply by \( (2a + 5b) \):** \[ (9a^{2} - b^{2})(2a) + (9a^{2} - b^{2})(5b) \] \[ = 18a^{3} - 2a b^{2} + 45a^{2}b - 5b^{3} \] **Arrange in standard form:** \[ 18a^{3} + 45a^{2}b - 2ab^{2} - 5b^{3} \] **Simplified Form:** \[ 18a^{3} + 45a^{2}b - 2ab^{2} - 5b^{3} \] --- **Summary of Simplified Expressions:** 1. \(-x^{2} - 6xy\) 2. \(34y^{2} + 16y - 7\) 3. \(-4x^{2} + 16xy - 19y^{2}\) 4. \(2x^{6} - 9y^{2}\) 5. \(18a^{3} + 45a^{2}b - 2ab^{2} - 5b^{3}\)

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Mind Expander

Let's tackle each expression one by one! 1. \( 2x(3x-4y) - (7x^2 - 2xy) \): First, distribute \(2x\) to get \(6x^2 - 8xy\). Now, simplify: \[ 6x^2 - 8xy - 7x^2 + 2xy = (6x^2 - 7x^2) + (-8xy + 2xy) = -x^2 - 6xy. \] 2. \( (5y+1)^2 - (3y+4)(2-3y) \): Start with the first part: \((5y+1)^2 = 25y^2 + 10y + 1\) and expand the second part: \((3y+4)(2-3y) = 6y - 9y^2 + 8 - 12y = -9y^2 - 6y + 8\). Putting these together gives: \[ (25y^2 + 10y + 1) - (-9y^2 - 6y + 8) = 25y^2 + 10y + 1 + 9y^2 + 6y - 8. \] Combine like terms: \(34y^2 + 16y - 7\). 3. \( (2x+y)^2 - (3x-2y)^2 + (x-4y)(x+4y) \): Start by expanding the squares: \((2x+y)^2 = 4x^2 + 4xy + y^2\) and \((3x-2y)^2 = 9x^2 - 12xy + 4y^2\). The difference yields: \[ (4x^2 + 4xy + y^2) - (9x^2 - 12xy + 4y^2) = -5x^2 + 16xy - 3y^2. \] Now for \((x-4y)(x+4y) = x^2 - 16y^2\). Combine: \(-5x^2 + 16xy - 3y^2 + x^2 - 16y^2 = -4x^2 + 16xy - 19y^2\). 4. \( x^6 + (x^3 - 3y)(x^3 + 3y) \): The second part is a difference of squares: \((x^3)^2 - (3y)^2 = x^6 - 9y^2\). Combine with the \(x^6\) term: \[ x^6 + x^6 - 9y^2 = 2x^6 - 9y^2. \] 5. \( (3a+b)(3a-b)(2a+5b) \): Recognizing the format, use the difference of squares: \((3a)^2 - b^2 = 9a^2 - b^2\). Now multiply with \((2a + 5b)\): \[ (9a^2 - b^2)(2a + 5b) = 18a^3 + 45a^2b - 2ab^2 - 5b^3. \] And there you have the expanded and simplified versions!

Related Questions

1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { (b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { 1) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b)\end{array} \)
Algebra South Africa Jan 22, 2025
\begin{tabular}{l} (a) Simplify \( \frac{2}{y^{0}} \) \\ (b) Simplify fully \( \left(16 a^{4}\right)^{\frac{3}{4}} \) \\ 5 \\ Factorise fully \( 18 c^{3} d^{2}-21 c^{2} \) \\ (Total for Question 4 is 3 marks) \\ \hline\end{tabular}
Algebra United Arab Emirates Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { \b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { \l) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b) \\ \text { 2. Find the products. } & \text { b) }(a+3)(a+4) \\ \text { a) }(x+1)(x+3) & \text { (d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { (e) }(3 x y+2 z)(2 x y-3 z) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { k) }(2 a-b)(2 a+b) & \text { n) }(x+3)^{2} \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \end{array} \)
Algebra South Africa Jan 22, 2025
1. Find the products. a) \( (x+2)(x+3) \) c) \( (a-2)(a+3) \) e) \( (2 y+5)(3 y-2) \) g) \( (2 a-3)(3 a-2) \) i) \( (x-3)(x-2) \) k) \( (2 x-3 y)(2 x+4 y) \) m) \( (7 x+3 y)(2 x-4 y) \)
Algebra South Africa Jan 22, 2025
An electrician charges an initial fee of \( \$ 50 \) and \( \$ 190 \) after 4 hours of worl a. Write a linear model that represents the total cost as a function of the \[ y=35 x+50 \] b. How much does the electrician charge per hour? \[ 4 x+x=140+5= \]
Algebra United States Jan 22, 2025

Latest Algebra Questions

2. Find the products. \( \begin{array}{ll}\text { a) }(x+1)(x+3) & \text { b) }(a+3)(a+4) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { e) }(3 x y+2 z)(2 x y-3 z) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { k) }(2 a-b)(2 a+b) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \text {-n) }(x+3)^{2}\end{array} \)
Algebra South Africa Jan 22, 2025
5. \( \begin{aligned} j-9 \pi= & 12 \pi \\ +9 \pi & +9 \pi\end{aligned} \)
Algebra United States Jan 22, 2025
7 Expand and simplify \( 5 x(3 x+4)(2 x-1) \)
Algebra United Arab Emirates Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { (b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { 1) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b)\end{array} \)
Algebra South Africa Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { \b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { \l) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b) \\ \text { 2. Find the products. } & \text { b) }(a+3)(a+4) \\ \text { a) }(x+1)(x+3) & \text { (d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { (e) }(3 x y+2 z)(2 x y-3 z) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { k) }(2 a-b)(2 a+b) & \text { n) }(x+3)^{2} \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \end{array} \)
Algebra South Africa Jan 22, 2025
Prev Next
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy