TRUE AND FALSE TYPE QUESTIONS 1 When 14x 3 – 3x 2 4x 2 is divided by 2x – 1, the remainder is 5 2 4x 2 – 8x 15 is exactly divisible by 2x – 1 3 x 2 – 5x 6 cannot be written as a product of two linear factors 4 x 3 – 3x 2 y 3xy 2 – y 3 is exactly divisible by x – y 5 (2a b) 2 – (2b a) 2 = 3(a 2 – b 2) is an identify 6 (x 2 1) 4 = (x 4 – 6x 2 1The perfect cube forms (x y) 3 (xy)^3 (x y) 3 and (x − y) 3 ( xy)^3 (x − y) 3 come up a lot in algebra We will go over how to expand them in the examples below, but you should also take some time to store these forms in memory, since you'll see them often Definition binomial A binomial is an algebraic expression containing 2 terms For example, (x y) is a binomial We sometimes need to expand binomials as follows (a b) 0 = 1(a b) 1 = a b(a b) 2 = a 2 2ab b 2(a b) 3 = a 3 3a 2 b 3ab 2 b 3(a b) 4 = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4(a b) 5 = a 5 5a 4 b 10a 3 b 2 10a 2 b 3 5ab 4 b 5Clearly, doing
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Expand (1/x+y/3)^3 class 9
Expand (1/x+y/3)^3 class 9-To find the tenth term, I plug x, 3, and 12 into the Binomial Theorem, using the number 10 – 1 = 9 as my counter 12C9 ( x) 12–9 (3) 9 = (2) x3 (196) = x3 Find the middle term in the expansion of (4x – y)8 Since this binomial is to the power 8, there will be nine terms in the expansion, which makes the fifth term the middle oneClass XICBSEMathematics Polynomials Practice more on Polynomials Page 2 wwwembibecom Hence the given expression x 10 y 3 t 50 is not a polynomial in one
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R S Aggarwal and V Aggarwal Solutions for Class 9 Mathematics CBSE, 3 Factorisation of Polynomials All the solutions of Factorisation of Polynomials Mathematics explained in detail by experts to help students prepare for their CBSE examsDesmos offers bestinclass calculators, digital math activities, and curriculum to help every student love math and love learning math Question 5 Find the remainder when x 3 x 2 x 1 is divided by x – using remainder theorem Solution Let p (x) = x 3 x 2 x 1 and q (x) = x – Here, p (x) is divided by q (x) ∴ By using remainder theorem, we have Question 6 Find the common factor in the quadratic polynomials x 2 8x 15 and x 2 3x – 10
If ` (x)/(y)(y)/(x)=1`, where` x,y ne 0` then the value of `(x^3y^3)` is A 1 B 1 C 0 D `(1)/(2)` class9; Ex 32, 11 If x 8(2@3) y 8(−1@1) = 8(10@5) , find values of x and y x 8(2@3) y 8(−1@1) = 8(10@5) 8(2𝑥@3𝑥) 8(−𝑦@𝑦An adfree subscription with Shaalaacom provides you an adfree experience across our site and app for your account The benefits of an adfree subscription will allow you to focus on what's important and also speeding up the experience by removing
They have the same value for all values of \ (xExample Expand 3 × (52) Answer It is now expanded We can also complete the calculation 3 × (52) = 3 × 5 3 × 2 = 15 6 = 21 In Algebra In Algebra putting two things next to each other usually means to multiply So 3(ab) means to multiply 3 by (ab) Here is an example of expanding, using variables a, b and c instead of numbersIn mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theoremCommonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written () It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 x) n, and is given by the formula =!!()!For example, the fourth power of 1 x is
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Transcript Ex 25, 6 Write the following cubes in expanded form (i) (2x 1)3 (2x 1)3 Using (a b)3 = a3 b3 3ab(a b) Where a = 2x & b =1 = (2x)3 (1)3 3(2x)(1) (2x 1) = 8x3 1 6x(2x 1) = 8x3 1 12x2 6x = 8x3 12x2 6x 1 Ex 25, 6 Write the following cubes in expanded form (ii) (2a 3b)3 (2a 3b)3 Using (x y)3 = x3 y3 3xy(x y) Where x = 2a & y = 3b = (2a)3Algebra Calculator is a calculator that gives stepbystep help on algebra problems See More Examples » x3=5 1/3 1/4 y=x^21 Disclaimer This calculator is not perfect Please use at your own risk, and please alert us if something isn't working Thank youML Aggarwal Solutions for Class 9 Maths Chapter 3 – Expansions are provided here to help students prepare and excel in their exams This chapter mainly deals with problems based on expansions Experts tutors have formulated the solutions in a step by step manner for students to grasp the concepts easily From the exam point of view, solving
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Worked examples 1 5y 15 = 5 × y 5 × 3 (HCF is 5) 2 21 x − 24 y = 3 × 7 x − 3 × 8 y (HCF is 3) = 5(y 3) = 3(7 x − 8 y) 3 12 ab 18 a = 6 a × b 6 aClick here👆to get an answer to your question ️ If x y = 12 and xy = 27 , find the value of x^3 y^3 eg {1, 2, 3, 6, 9, 18} is the set of factors of 18 expand factorise 5a(a − 2) 5a2 − 10a 'Gnidnapxe' is the reverse of 'expanding' It's 'factorising', you dummy!
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Write expression log(x4y7 z16) log ( x 4 y 7 z 16) as a sum or difference of logarithms with no exponents Simplify your answer completely log(x4y7 z16) = log ( x 4 y 7 z 16) = Get help Box 1 Enter your answer as an expression Example 3x^21, x/5, (ab)/c Be sure your variables match those in the questionWe can skip n=0 and 1, so next is the third row of pascal's triangle 1 2 1 for n = 2 the x^2 term is the rightmost one here so we'll get 1 times the first term to the 0 power times the second term squared or 1*1^0* (x/5)^2 = x^2/25 so not here 1 3 3 1 for n = 3 Squared term is second from the right, so we get 3*1^1* (x/5)^2 = 3x^2/25 so not(),where f (n) (a) denotes the n th derivative of f evaluated at the point a
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Suppose we want to expand (2xy)3 We pick the coefficients in the expansion from the relevant row of Pascal's triangle (1,3,3,1) As we move through the terms in the expansion from left to right we remember to decrease the power of 2x and increase the power of y So, (2xy)3 = 1(2x)3 3(2x)2y 3(2x)1y2 1y3 = 8x3 12x 2y 6xy y3 ExampleFactorisation 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;Find (3 x) 3 The power that we are expanding the bracket to is 3, so we look at the third line of Pascal's triangle, which is 1 3 3 1 So the answer is 3 3 3 × (3 2 × x) 3 × (x 2 × 3) x 3 (we are replacing a by 3 and b by x in the expansion of (a b) 3 above) Generally
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