Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreFactorise 3x5 – 6x4 – 2x3 4x2 x – 2 asked May 21, 19 in Class VIII Maths by priya12 (12,625 points) factorizationFactorise 5 x 2 y − 1 5 x
X Y Z Whole Square Formula A Pictures Of Hole 18
Factorise x^4-y^4+x^2-y^2
Factorise x^4-y^4+x^2-y^2-Factorise 49(X Y)2 9(2x Y)2 CISCE ICSE Class 8 Textbook Solutions 6351 Important Solutions 5 Question Bank Solutions 7027 Concept Notes & Videos 1 Syllabus Advertisement Remove all ads Factorise 49(X Y)2 9(2x Y)2 Mathematics Sum Factorise 49(xFactorise 25(2x Y)2 16(X Y)2 CISCE ICSE Class 8 Textbook Solutions 6351 Important Solutions 6 Question Bank Solutions 7107 Concept Notes & Videos 196 Syllabus Advertisement Remove all ads Factorise 25(2x Y)2 16(X Y)2 Mathematics Sum Factorise
If the polynomial k 2 x 3 − kx 2 3kx k is exactly divisible by (x3) then the positive value of k is ____;SolutionShow Solution 75 (x y) 2 48 (x y) 2 = 3 25 (x y) 2 16 (x y) 2 = 3 {5 (x y) 2 } {4 (x y)} 2 Using a 2 – b 2 = (a b) (a – b) = 3 5 (x y) 4 (x y) 5 (x y) 4 (x y) = 3 5x 5y 4x 4y 5x 5y 4x 4y = 3 (9x y) (x 9y) · Ex 141, 3 Factorise (i) 𝑥^2 𝑥𝑦 8𝑥 8𝑦𝑥^2 𝑥𝑦 8𝑥 8𝑦 = (𝑥^2 𝑥𝑦) (8𝑥 8𝑦)Both have x as common factorBoth have 8 as common factor = 𝑥 (𝑥𝑦)8 (𝑥𝑦)Taking (𝑥𝑦) common = (𝒙𝒚) (𝒙 8)
· It can be factored as x^2 y^2 = (xy)(xy) Notice that when you multiply (xy) by (xy) then the terms in xy cancel out, leaving x^2y^2 (xy)(xy) = x^2xyyxy^2 = x^2xyxyy^2 = x^2y^2 In general, if you spot something in the form a^2b^2 then it can be factored as (ab)(ab) For example 9x^216y^2 = (3x)^2(4y)^2 = (3x4y)(3x4y)This is true of the "difference of squares" you sent us, x 2 y 2 Once you think you know the factors you can check by multiplication Here are three possibilities for factorizations of x 2 y 2 (x y)(x y) (x y)(x y) (x y)(x y) Expand each of them and see which simplifies to x 2 y 2Factor Consider x^ {2}y^ {2}xy22xy as a polynomial over variable x Consider x 2 y 2 x y − 2 2 x y as a polynomial over variable x Find one factor of the form x^ {k}m, where x^ {k} divides the monomial with the highest power x^ {2} and m divides the constant factor y^ {2}y2 One such factor is xy1
Factorise (xy) 2 7(x 2y 2)12(xy) 2 Share with your friends Share 2 Well done Thunder Sam Keep it up!!6 ; · So, in your case, a = x y, b = 8 And the factorisation is (x y 8)(x y 8) 1 1 frak1a Lv 6 1 decade ago It is the difference of 2 squares 0 1 Still have questions?Factorise 75(x y)2 48(x y)2 The sum of first n natural numbers is given by the expression n^2 /2 n/2
To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Factorise `x^2xyxzyz`Factorise 1 3x 3y 4(x y)2 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queries1711 · Therefore, by using the identity (xy) 2 = x 22xyy 2 p 2 –10p25 = (p5) 2 (iii) 25m 2 30m9 Ans Given 25m 2 30m9 Since, 25m 2 , 30m and 9 can be substituted by (5m) 2, 2×5m×3 and 3 2 respectively we get, = (5m) 2 2×5m×3 3 2 Therefore, by using the identity (xy) 2 = x 2 2xyy 2 25m 2 30m9 = (5m3) 2 (iv) 49y 2 84yz36z 2 Ans Given 49y 2 84yz36z 2
Divide y, the coefficient of the x term, by 2 to get \frac{y}{2} Then add the square of \frac{y}{2} to both sides of the equation This step makes the left hand side of the equation a perfect square∵ x4 (y z)4 = x22 (y z)22 = (x2) (y z)2 (x2) (y z)2 Using a2 b2 = (ab)(ab)We can factorise x2 (y z)2 further as x2 (y z)2Since x^2y^2=\frac12\left((xy)^2(xy)^2\right) the minimum comes when xy is smallest, that is 1 if xy is odd Thus, the minimum is \frac12\left((xy)^21\right) Since x 2 y 2 = 2 1 ( ( x y ) 2 ( x − y ) 2 ) the minimum comes when ∣ x − y ∣ is smallest, that is 1 if x y is odd
· x^2xy y^2 this equation can't be factor because there is no two numbers whose product and sum is oneFactorise the following (i) 4x^2 9y^2 25z^2 12xy 30yz xz (ii) 25x^2 4y^2 9z^2 – xy 12yz – 30xz asked Oct 30, in Algebra by Darshee ( 491k points) algebra · Ex 141, 2 (Method 1) Factorise the following expressions (x) a𝑥^2y bx𝑦^2 cxyza𝑥^2y = a × 𝑥^2 × y = a × 𝑥 × 𝑥 × yb𝑥𝑦^2 = b × 𝑥 × 𝑦^2 = b × 𝑥 × y × yc𝑥𝑦𝑧 = c × 𝑥 × y × zSo, x and y are the common factorsa𝑥^2y b〖𝑥𝑦〗^2 c𝑥𝑦𝑧 = (a × 𝑥 × 𝑥 × y) (b × 𝑥 × y × y) (c × 𝑥 × y
0118 · Explanation Making m = x −y n = y −z p = x −z we have m2 n2 p2 = 2(x2 y2 z2 − x ⋅ y − x ⋅ z −y ⋅ z) then x2 y2 z2 −x ⋅ y − x ⋅ z − y ⋅ z = 1 2((x − y)2 (y − z)2 (x −z)2)Get your answers by asking now Ask Question 100 Join Yahoo Answers and get 100 points today Join · Read as factorize x y whole squared 2(xy) Yahoo Answers is shutting down on May 4th, 21 (Eastern Time) and beginning April th, 21 (Eastern Time) the Yahoo Answers website will be in readonly mode
Factorise a2b – b3 Using this result find the value of 1012 x 100 – 1003 asked May 23, 19 in Class VIII Maths by muskan15 ( 3,440 points) factorizationClick here👆to get an answer to your question ️ Factorise xy(z^2 1) z(x^2 y^2)Please see the explanation x(x y)^2 xy(x y) =x(xy)(xy y) =x^2(xy) (x y)^2 10(x y)25 Using a^22abb^2=(ab)^2 =(xy) 5^2=> Or =(xy5)^2 x^2
Share It On Facebook Twitter Email 1 Answer 1 vote answered May 15, by Subnam01 (5k points) selected May 16, by Puskar Best answer 15x 3 y 2 z = 3 × 5 × x × x × x × y × y × zYou factor as follows Put a = (xy) Then 3(xy)^2 2(xy) becomes 3a^2 2a which is very easy to factor since a divides into each term Clearly 3a^2–2a =Case 2 xy1=(xy1) Lets take Case 1 xy1=xy1 => 2=0 This means that if we assume that there exists a value of (xy) which satisfies the Case 1 equation, then 2 and 0 would be equal Since that is absurd, our initial assumption was wrong So, there's no common answer to your equation and the Case 1 equation For Case 2 xy1=(xy1) => 2x2y=0 => xy=0
Factorise fully 4(x y)^2 9(x y)^2 2 See answers aleighablock878 aleighablock878 Answer ( 5x y ) ( x 5y ) or 5x^2 5y^2 26xy irspow irspow Answer Stepbystep explanation 4(x^22xyy^2)9(x^22xyy^2) 4x^28xy4y^29x^218xy9y^25x^226xy5y^25x^2xy25xy5y^2x(5xy)5y(5xy) (x5y)(5xy) or you can also remove theClick here👆to get an answer to your question ️ Factorise x^2(y z) y^2 (z x) z^2(x y)To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Factorise `x^4x^2y^2y^4`
What must be subtracted from 4x^42x^36x^22x6 so that the result is exactly divisible by 2x^2x1?Here is your solution☺ =============================== To factorise x² y²/100 => x² ( y/10)² => ( x y/10 ) ( x y/ 10) ♦ Note a² b² = (a b) (a b) here this identity is used =============================== Thanks☺Factorise {eq}\displaystyle (x y)^2 9 (x y) {/eq} Factorization when multiplied the product of the polynomial expression together they give the polynomial expression again is the
View Full Answer (xy) 27(x 2y 2)12(xy) 2 now by using a 2b 2 =(ab)(ab), we have (xy) 27(xy)(xy)12(xy) 2 we take xy = z and xy =a now we have, z 27za12a 2 = z 23za4za12a 2 = z(z3)4a · x^2 2xy y^2 = (x y)(x y) = (x y)^2"y is a root" simply means that the only way we can get the thing to add up to zero, is to make x = y for example, if x = 3 and y = 3, then we get x^2 2xy y^2 = 9 2(3)(3) 9 = 9 18 9 = 0 or (x y)^2 = (33)^2 = 0^2 = 0However, our factorization will work for any value of x and yYou can observe that $xy = (xy)$ So both $x^2y^2$ and $xy$ have $xy$ as a factor You can use these info to get below factorization $$ x^2 y^2 x y = (xy)(xy)(xy) = (xy1)(xy
Factor x^2y^2 x2 − y2 x 2 y 2 Since both terms are perfect squares, factor using the difference of squares formula, a2 −b2 = (ab)(a−b) a 2 b 2 = ( a b) ( aClick here👆to get an answer to your question ️ Factorise (x^2 2xy y^2) z^2Factorise 15x 3 y 2 z 25 x y 2 z 3 cbse;
Factorise `x^32x^2x2` Factorise `x^32x^2x2` Watch later Share Copy link Info Shopping Tap to unmute If playback doesn't begin shortly, try restarting your deviceClick here👆to get an answer to your question ️ Factorise 2(x y)^2 9(x y) 52x27xy5y2 Final result (5y 2x) • (x y) Step by step solution Step 1 Equation at the end of step 1 ((0 (2 • (x2))) 7xy) 5y2 Step 2 Equation at the end of step 2 ((0 What is the degree of \displaystyle{9}{x}{y}^{{3}}{z}^{{6}} ?
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