Factorisation is the opposite process of expanding brackets For example, expanding brackets would require \ (2 (x 1)\) to be written as \ (2x 2\) Factorisation would be to start with \ (2x 2\) and end up with \ (2 (x 1)\) The two expressions \ (2 (x 1)\) and \ (2x2\) are equivalent; Factorise a 12 x 4 a 4 x 12 Share with your friends Share 7 a 12 x 4 – a 4 x 12 = a 4 x 4 (a 8We have to factorise x 4y 4 Solution x4 – y 4 cxn ye expressed xs (x 2) 2 – (y2) 2 —(i) We know the identity \(x^{2}y^{2}= (xy)(xy)\) Hence using the
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X^4-16y^4 factorise-Factorise x^24x4 Updated On 9 To keep watching this video solution for FREE, Download our App Join the 2 Crores Student community now! Factorise x4 x2 1 Factorise x4 x2 1 how_to_reg Follow thumb_up Like (11) visibility Views (714K) edit Answer question_answer Answers(2) edit Answer person Kishore Kumar Consider x 4 x 2 1 = (x 4 2x 2 1) – x 2 = (x 2) 2 2x 2 1 – x 2 = x 2 1 2 – x 2 It is in the form of (a 2 – b 2) = (a b)(a
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Algebraic Methods wwwnaikermathscom 5 f(x) = 3x3 – 5x2 – 16x 12 (a) Find the remainder when f(x) is divided by (x – 2)(2) Given that (x 2) is a factor of f(x), (a) factorise f(x) completely(4) June 07 Q2 6 f(x) = 2x3 – 3x2 – 39x (a) Use the factor theorem to show that (x 4) is a factor of f (x)(2) (b) Factorise f (x) completely I am trying to factor x 4 1 in to two multiplied polynomials Homework Equations My teacher gave us this hint that its factored form is (ax2bxc)(ax2bxc) The Attempt at a Solution First i assumed that a and c were equal to 1 so that when x 2 is multiplied by the other x 2 is gives me x 4 and 1 times 1 gives me 1This is the same as x 2 0x 4, so you need to find factors of 4 that add up to 0
This is similar to other factorisation problems such as x 2 5x 6 In this problem, you would find prime factors of 6 that add up to 5 (In this case 3, 2) Now, do the exact same with this problem!Factorisation Factorisation class 9 Factorise x^8y^8 Factorise x^4y^4 Factorise x^4y^4 Factorise x^8y^8 Factorise x^2y^2#factorisationofpolynomiaQuotient of x^38x^217x6 with x3;
Add to your resource collection Ref R19 Printable/supporting materials Fullscreen mode Teacher notes Question Solution Question Show that (x2 2y2)2 −4x2y2 =x4 4y4, ( x 2 2 y 2) 2 − 4 x 2 y 2 = x 4 4 y 4, and hence factorise x4 4y4 x 4 4 y 4Factor x^41 x4 − 1 x 4 1 Rewrite x4 x 4 as (x2)2 ( x 2) 2 (x2)2 −1 ( x 2) 2 1 Rewrite 1 1 as 12 1 2 (x2)2 −12 ( x 2) 2 1 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) ( a b) where a = x2 a = x 2 and b = 1 b = 1 (x2 1)(x2 −1) ( x 2View more examples » Access instant learning tools Get immediate feedback and guidance with stepbystep solutions and Wolfram
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Factorise `(a) x^(4)4x^(2)16 " " (ii) x^(4)4`X 4 − (y z) 4 (x 2) 2 − ((y z) 1) 2 (x 2 − (y z) 2) (x 2 (y x) 2)X^4 324 (x^2)^2 18^2 That would be a perfect square if the term that is twice the product of the square roots of those two terms were added That's 2·18·x^2 or 36x^2 So we add that between those terms and then subtract it That's the same as adding 0 (x^2)^2 36x^2 18^2 36x^2 Factor the first three terms as (x^2 18) (x^2 18) or
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They have the same value for all values of \ (x 3 Let s = x y 2 so x, s, y make an arithmetic progression with difference name it d So x = s − d and y = s d Now E = ( s − d) 4 ( s d) 4 16 s 4 = 2 s 4 12 s 2 d 2 2 d 4 16 s 4 = 18 s 4 12 s 2 d 2 2 d 4 = 2 ( 9 s 4 6 s 2 d 2 d 4) = 2 ( 3 s 2 d 2) 2 Share answered May 1 ' at 1549 Answer3x^2 4x 3x 4x ( 3x 4 ) 1 ( 3x 4 )( 3x 4 ) ( x 1 )hope it helps
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Factorise 25/4 x^2y^2/9 Updated On To keep watching this video solution for FREE, Download our App Join the 2 Crores Student community now! Yes, the four solutions to x^44=0 are x=±1±i So to factor this completely, you have x^44=(x1i)(x1i)(x1i)(x1i) If you multiply the first two factors and the second two factors, you obtain x^44=(x^22x2)(x^2–2x2) Alternatively, if you multiply the first and last factor, and also the second and third, you get x^44=(x^2–2iWhich is what we wanted and hence factorise x4 4y4 x 4 4 y 4 We know that x4 4y4 =(x2 2y2)2 −4x2y2 x 4 4 y 4 = ( x 2 2 y 2) 2 − 4 x 2 y 2 Notice that the right hand side is of the form a2 −b2 a 2 − b 2, where a = x2 2y2 a = x 2 2 y 2 and b =2xy b = 2 x y
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Factorise x 4 /4 4/x 4 1 Share with your friends Share 3 x 4 4 4 x 4Try typing these expressions into the calculator, click the blue arrow, and select "Factor" to see a demonstration Or, use these as a template to create and solve your own problems Problem 4x2 −9 4 x 2 − 9 Solution (2x3)(2x −3) ( 2 x 3) ( 2 x − 3) Problem x4 −81 x 4 − 81Factoriser Factoriser Enter a quadratic equation to factorise For indices use (^) Factorise Submit
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Answer to Factorise \\text{x}^{4} \\text5{x}^{2} 6 By signing up, you'll get thousands of stepbystep solutions to your homework questions YouFor example putting 2x² x 3 into the form (2x 3)(x 1) factorise 9x to the power of 4 64y to the power of 2 Math Factorise X'29X 14 maths factorise a^7ab^6 You can view more similar questions or ask a new question x^4(xz)^4 can be factorised as, =x^4(xz)^4 =(x^2)^2(xz)^2^2 =x^2(xz)^2x^2(xz)^2 we used the following formula, a^2b^2=(ab)(ab) x(xz)x(xz)x^2x^2z^22xz we used the following formula, a^2b^2=(ab)(ab) (ab)^2=a^2b^22ab =x(xz)x(xz)x^2x^2z^22xz =(xxz)(2xz)2x^2z^22xz =z(2xz)2x^2z^22xz
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Factorise x^4x^21 Get the answer to this question and access a vast question bank that is tailored for studentsExpand polynomial (x3)(x^35x2) GCD of x^42x^39x^246x16 with x^48x^325x^246x16; Ex 142, 4 Factorise (iv) 𝑥^4 – 〖"(x – z)" 〗^4 𝑥^4 – 〖"(x – z)" 〗^4 = 〖(𝑥^2)〗^2 − (𝑥 − z)^2 ^2 Using 𝑎^2 − 𝑏^2 = (a b) (a − b) Here 𝑎=𝑥^2 and b = (𝑥 − z)^2 = 𝑥^2(𝑥 − z)^2 𝑥^2 −(𝑥 − z)^2 Asking Using 𝑎^2 − 𝑏^2 = (a b) (a − b) Here 𝑎=𝑥 and b = (𝑥 − z) Asking Using 𝑎^2 − 𝑏^
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Check x 4 is the square of x 2 Check y 4 is the square of y 2 Factorization is (x 2 y 2) • (x 2 y 2) Trying to factor as a Difference of Squares 12 Factoring x 2 y 2 Check x 2 is the square of x 1 Check y 2 is the square of y 1 Factorise Completelyx 4 – 2x 2 y 2 y 4 Answers x 4 – x 2 y 2 – x 2 y 2 y 4 x 2 (x 2 – y 2) – y 2 (x 2 – y 2) (x 2 – y 2) (x 2 – y 2) (x y) (x – y) (x y) (x – y) Kavungya answered the question on at 1045 Next Janice, a fruit vendor obtained a total of Kshs 6144 from her sales of Expression #=x^4y^4# Recall the factorization of the difference of two squares #a^2b^2 = (ab)(ab)# In our example, we will use this factorization twice Note #x^4 =(x^2)^2 and y^4 =(y^2)^2 # Applying the factorization above Expression #= (x^2y^2)(x^2y^2)# Now, the second factor above is also the difference of two squares
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Share It On Facebook Twitter Email 1 Answer 1 vote answered by Mohini01 (677k points) selected by Sunil01 Best answer x 4 – y 4 = (x 2) 2 – (y 2) 2 Identity a 2 – b 2 = (a b)(a – b)Here again we add and subtract suitable terms to bring it in a factorisation form Notice that x 4 4 y 4 = (x 2) 2 (2 y 2) 2 − 2 (x 2) (2 y 2) − (2 x y) 2 = (x 2 2 y 2) 2 − (2 x y) 2 The second expression is in the form a 2 − b 2, where a = (x 2 2 y 2) and b = 2 x y Hence we may write x 4 4 y 4 = (x 2 2 y 2 2 x y) (x 2 2 y 2 − 2 x y)Remainder of x^32x^25x7 divided by x3;
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SolutionShow Solution `x^4 y^4 27x^2y^2` = ` (x^2)^2 (y^2)^2 2x^2y^2 25x^2y^2` = ` (x^2 y^2)^2 25x^2y^2` = ` ( x^2 y^2 )^2 (5xy)^2` ∵ a2 b2 = ( a b ) ( a b ) = ` (x^2 y^2 ) 5xy ( x^2 y^2 ) 5xy` = ` x^2 5xy y^2 x^2 5xy y^2 `Watch Video in App This browser does not support the video element 1603 k 1190 k Answer Step by stepSo x^4 7x^2 8 = y^27y8 y^27y8 = (y8) (y1) substitute x back in (x^2 8) (x^21) Now to simplify this, I don't think you can factorise the first bracket with whole numbers but you can simplify the second bracket (x^21) = (x1) (x1) so x^4 7x^2 8 = (x^2 8) (x1) (x
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how do you factorise?Example 1 with x coefficient and a constant Fully factorise x2 6x 5 x 2 6 x 5 Write out the factor pairs of the last number (5) in order x2 6x 5 x 2 6 x 5 Factors of 5 1, 5 2 We need a pair of factors that to give the middle number ( 6) and to give the last number ( 5 ) If you can't find rational roots with the Rational Root Theorem, your next bet for solving this on paper is to factor the quartic into two quadratic equations For example, in this case, we can let 64 x 4 64 x 3 − x 2 − 51 x 39 = ( a x 2 b x c) ( p x 2 q x r) Note that, by comparing coefficients, we can tell that a p = 64, c r = 39,
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Transcript Ex 24, 4 Factorise 12x2 7x 1 12x2 7x 1 We factorize using the splitting the middle term method = 12x2 4x 3x 1 = 4x (3x 1) 1 (3x 1) = (3x 1) (4x 1) Ex 24, 4 Factorise (ii) 2x2 7x 3 2x2 7x 3 We factorize using the splitting the middle term method = 2x2 6x x 3 = 2x (x 3) 1 (x 3) = (x 3) (2x 1) Ex 24, 4 Factorise (iii) 6x2 5x 6 6x2 5x 6 WeFactor x^24 x2 − 4 x 2 4 Rewrite 4 4 as 22 2 2 x2 − 22 x 2 2 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) ( a b) where a = x a = x and b = 2 b = 2 (x2)(x− 2) ( x 2) ( x 2)Factor x 4 324 (x 2) 2 18 2 That would be a perfect square if the term that is twice the product of the square roots of those two terms were added
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X 4 coefficient is 1 we know the coefficient for the x 2 terms in the factor will also be 1 (x 2) (x 2) Because the constant is a negative and the coefficient for the x term is a negative we know the sign for the constants in the factors will have one positive and one negative (x 2 ) (x 2 −) Now we need to determine the Factorise x 4 – y 4 factorisation;X 4 y 4 3x 2 y 2 = x 4 y 4 2x 2 y 2 x 2 y 2 = (x 2) 2 (y 2) 2 2x 2 y 2 x 2 y 2 = ( x 2 y 2) 2 (xy) 2 = ( x 2 y 2 xy )( x 2 y 2 xy ) ∵ a 2 b 2 = ( a b )( a b )
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Math\text{There are multiple ways to approach this problem}/math Approach # 1 mathx^4 x^2 1 = x^4 x^3 x^2 x^3 x^2 x x^2 x 1 /math mathQuestion what is the process to factorise x^4 324 Answer by Edwin McCravy() (Show Source) You can put this solution on YOUR website!Factorisation Zigya App Factorise x 4 (y z) 4 ∵ x 4 (y z) 4 = x 2 2 (y z) 2 2 = (x 2) (y z) 2 (x 2) (y z) 2 Using a 2 b 2 = (ab) (ab) We can factorise x 2 (y z) 2 further as x 2 (y z) 2 = (x) (y z) (x) (y z)
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Get answer Factorise (a) x^(4)4x^(2)16 " " (ii) x^(4)4 Apne doubts clear karein ab Whatsapp par bhi Try it nowWatch Video in App This browser does not support the video element 600 100 AnswerFactor x^45x^24 full pad » x^2 x^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot \msquare {\square} \le \ge
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