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Solution. Lydia. Mechanical engineer · Tutor for 4 years. Answer: 5 (x+2y^ {2}) (x^ {2}-2xy^ {2}-4y^ {4}) 5(x+2y2)(x2−2xy2 −4y4) . Explanation: Factor the expression: 5 (x^ {3}-8y^ {0}) 5(x3−8y0) Rewrite the expression into the form of a^ {3}1b^ {3}-b^ {3} a31b3−b3 I 5 (x^ {3}- (2y^ {2})^ {3}) 5(x3−(2y2)3) Factor the expression using ...
18 maj 2015 · You can use a formula for a sum of cubes: a3 +b3 = (a +b) ⋅ (a2 − ab +b2) Notice that 40y6 = 5 ⋅ 8 ⋅ y6 = 5 ⋅ 23 ⋅ (y2)3 = 5 ⋅ (2y2)3. Using the above formula for a = x and b = 2y2 and factoring out constant 5, we get: 5x3 + 40y6 = 5x3 +5(2y2)3 = 5(x + 2y2)(x2 − 2xy2 +4y4)) Answer link.
What is factor 5x^3+40y^6 ? The solution to factor 5x^3+40y^6 is 5(x+2y^2)(x^2-2xy^2+4y^4)
Solution - Factoring binomials using the difference of squares. 5 ⋅ ( x + 2 y 2) ⋅ ( x 2 − 2 x y 2 + 4 y 4) Step by Step Solution. Step 1 : Equation at the end of step 1 : (5 • (x3)) + (23•5y6) Step 2 : Equation at the end of step 2 : 5x3 + (23•5y6) Step 3 : Step 4 : Pulling out like terms : 4.1 Pull out like factors :
