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Two blocks of masses m_(1) = 10 kg and m_(2) = kg are connected by a spring of stiffness k = 0 N/m The coefficient of friction between the blocks and the fixed horizontal surface isPq p q xx abc ab a b x x x pq qr rp x y x y x y z Difficulty level Easy What is known Terms Factorise the following expressions 2 3 2 22 2 2 2 2 2 2 2 22 (i) 7 42 (ii) 6 12 (iii) 7 14 (iv) 16 16 2 2 2 2 2 2 5 zz z z z z u u u u u u u u u The common factors are 2, 2, and z 3 2(m) 36 d2 (n) 4 q2 (o) 49 w2 (p) x2 64 9 Factorise the following expressions, which contain a difference of squares (a) a2 4b2 (b) x2 25y2 (c) p2 64q2 (d) 16c2 d2 (e) 81 4g2 (f) 36w2 y2 (g) 4a2 1 (h) g2 81h2 (i) 49x2 y2 (j) 9c2 16d2 (k) 4p2 9q2 (l) b2 100c2 (m) 25 16a2 (n) 4d2 121 (o) 225 49k2 (p) 9x2 0·25 10 Factorise the following

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Factorise (p+q)^2-20(p+q)-125

Factorise (p+q)^2-20(p+q)-125-Factorise (i) \( 4 p^{2}=9 q^{2} \) (ii) \( 63 a^{2}112 b^{2} \) (iii) \( 49 x^{2}36 \) (iv) \( 16 x^{3}144 x \) (v) \( (lm)^{2}(lm)^{t} \) (vi) \(9 x^{2} y^{2Factorise the following a 2 – 16p – 80 Advertisement Remove all ads Solution Show Solution We have, `p^2 16p 80 = p^2 ( 4)p 80` = `p^2 p 4p 80` = `p(p ) 4(p )` = `(p )(p 4)` Concept Expansion of (a b)2 = a2 2ab b2

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The topics that are covered in NCERT Solutions for Class 8 Maths Chapter 9 are 91 – introduction to expressions, 92 – terms, factors and coefficients, 93 – monomials, binomials and polynomials, 94 – like and unlike terms, 95 – addition and subtraction of algebraic expressions, 96 – multiplication of algebraic expressionsGiven that, p, q, p,q, p, q, and r r r are its roots, (x − p) (x − q) (x − r) = 0 (xp)(xq)(xr)=0 (x − p) (x − q) (x − r) = 0 is the required cubic equation Since it can be represented in the form a x 3 b x 2 c x d = 0 x 3 b a x 2 c a x d a = 0 ax^3bx^2cxd=0 \implies x^3 \frac{b}{a} x^2 \frac{c}{a} x \frac{dHow would you factorise this (pq)^25(pq) to get this answer(pq)(pq5) math Cn someone factorise 3a^221a24 Mathematics Please help me factorise 8m24mn Algebra Factorise 6x^3 11x^2 3x math factorise 15x 6x5y2 math help please!

 x^39x^224x = (x2)(x2)(x5) f(x) = x^39x^224x By the rational root theorem, any rational zeros of f(x) are expressible in the form p/q for integers p, q with p a divisor of the constant term and q a divisor of the coefficient 1 of the leading term Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 8 so that you can refer them as and when required The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions For a better understanding of this chapter, you should also see summary of Chapter 14(2x1)P(x) = 4x^2 4x 3 now taking the multiplied factor of 2x1 from LHS to RHS and dividing the quadratic with the linear ex

Step 1 Learn (4 videos) The first in a series of videos on indices and surds In this video we look at laws of indices at a Year 1 A level standard The second in a series on indices and surds This video extends the use of index laws to include negative and fractional indicesThus solving P(x) = 0 is reduced to the simpler problems of solving Q(x) = 0 and R(x) = 0 Conversely, the factor theorem asserts that, if r is a root of P(x) = 0, then P(x) may be factored as = (), where Q(x) is the quotient of Euclidean division of P(x) = 0 by the linear (degree one) factor x – r

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The common factors are 2, y The common factors are 2, 7, p, q The common factor is 1 12 a2b = 2 × 2 × 3 × a × a × b The common factors are 2, 3, a, b −4 x2 = −1 × 2 × 2 × x × x The common factors are 2, 2, x The common factors are 2, 5 (viii) 3 x2y3 = 3 × x × x × y × y × yAnswer (1 of 3) (2x1)P(x) = (2x^2 5x 2)(6x^2 x 1) We have just opened the brackets and simplified the expression then taken it to RHS simplifying LHS;(i) ax 2 bx (ii) 7p 2 21q 2 (iii) 2x 3 2xy 2 2xz 2 (iv) am 2 bm 2 bn 2 an 2 (v) (lm l) m 1 (vi) y(y z) 9( y z) (vii) 5y 2 y 8z 2yz (viii) 10ab 4a 5b 2 (ix) 6xy 4y 6 9x Solution For part (i), (ii), (iii), and (vi) First we will find factors of each term then find out which factors are common in each term, and take out that common factor from

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 যদি , তারপরে P QR সন্ধান করুন, যেখানে P, Q এবং R পুরো সংখ্যা whole 869kAnswer (i) 12x = 2 x 2 x 3 x x 36 = 2 x 2 x 3 x 3 The common factors are 2, 2, 3 And, 2 x 2 x 3 = 12 (ii) 2y = 2 x y 22xy = 2 x 11 x x x y The common factors are 2Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more

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Vedantu’s NCERT Solutions for Class 8 maths chapter 14 is a good platform for the students to explore more knowledge and can gain command over the chapter All the concepts are explained in your stepbystep process with plenty of examples The test of papers provided in our solutions is good enough to make the students get to practiceThis is what we shall do now 1421 Method of common factors • We begin with a simple example Factorise 2x 4 We shall write each term as a product of irreducible factors; Exercise 141 Question 1 Find the common factors of the terms (i) 12 x , 36 (ii) 2 y , 22 xy (iii) 14 pq , 28 p 2 q 2 (iv)

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Click here👆to get an answer to your question ️ Factorize the following expressions 25x^2 4y^2 9z^2 xy 12 yz 30 zxFactorise 5(3x4y)^2(2xy)^2 Created by kukudevangi16 Math BRAINLYSAMI Answer 35x^260y^xy Stepbystep explanation 45x^280y^21xy80x^2y^280xy10 p 2 − q pq − p 11 m 3 − m 2 − mn m n − 1 12 r 2 ap − ar − rp 13 mn − ma − an a 2 bn − ba 14 1 p pq p 2 q 15 4a 4b − 5cx − 5bx 4c − 5ax 16 4x 5 − 8x 4 − 5x 3 10x 2 x − 2 17 y − 1 − (y − 1) 2 4 − 4y 18 36x 2 − 16y 2 19 64 − 32x 4x 2 a2 10a 25

Doppler Radar Analysis Of The Northfield Texas Tornado Of 25 May 1994 In Monthly Weather Review Volume 125 Issue 2 1997

Doppler Radar Analysis Of The Northfield Texas Tornado Of 25 May 1994 In Monthly Weather Review Volume 125 Issue 2 1997

Answer (i) p2 6 p 8 It can be observed that, 8 = 4 × 2 and 4 2 = 6 ∴ p2 6 p 8 = p2 2 p 4 p 8 = p ( p 2) 4 ( p 2) = ( p 2) ( p 4) (ii) q2 − 10 q 21 It can be observed that, 21 = (−7) × (−3) and (−7) (−3) = − 10 ∴ q2 − 10 q 21 = q2 − 7 q − 3 q 21 (v) p 3 q 6 − p 6 q 3 = p 3 q 3 (q 3 − p 3) (p 3 q 6 – p 6 q 3) ÷ p 3 q 3 = p 3 q 3 q 3 – p 3 p 3 q 3 = q 3 – p 3 P3 Work out the following divisions (i) (10x − 25) ÷ 5 (ii) (10x − 25) ÷ (2x − 5) (iii) 10y(6y 21) ÷ 5(2y 7) (iv) 9x 2 y 2 (3z − 24) ÷ 27xy(z −Answer to 5 Factorise each expression a 9a 21b b 4c 18d c 12p2 q d 35 14man e 25y2 15x f 16a2 b g 42m2 12n h 63p2 81 27y i

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NCERT solutions for Class 8 Maths Chapter 14 FactorisationExercise 141 Page 2 Find the common factors of the terms Find the common factors of the terms 16x 3, −4x 2, 32x Find the common factors of the terms 10 pq, qr, 30 rp Find the common factors of the terms 3x 2 y 3, 10x 3 y 2, 6x 2 y 2 z Find the common factors of the terms 7Q3 (a) Simplify n 0 (1) (b) Simplify (3 x 2 y 5) 3 (2) (c) Factorise fully 2 e 2 – 18 (2) (d) Make r the subject of 𝑚=√ 6𝑎𝑟 5𝑟 (4) (Total for question = 9 marks)Hence P 3 Q3d 8R3d2 6PQRd = f 1 P 3 Q3d R 3 d 2 3PQRd = 1 9Rd(r2d P Q ) = Hence 0 = a Then either R = 0 and P Q V ~ i i s binomial unit and so n possibility from Lemma 2, or R2d PQ = 0 Then 1 is the only I/?

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2x =2× x 4 =2 × 2 Hence 2x 4 = (2 × x) (2 × 2) Notice that factor 2 is common to both the terms (i) 7x 42 (ii) 6p 12q (iii) 7a 2 14a (iv) 16z z 3 (v) l 2 m 30alm (vi) 5x 2 y 15xy 2 (vii) 10a 215b 2 c 2 (viii) 4a 2 4ab 4ca (ix) x 2 yz xy 2 z xyz 2 (x) ax 2 y bxy 2 cxyz Solution Factorization is a method of finding factors for any mathematical object, be it a number, a polynomial or any algebraic expression Thus, factorization of an algebraic expression Factorise the following expressions (i) 7x − 42 (ii) 6p − 12q (iii) 7a 2 14a (iv) −16z z 3 (v) l 2 m 30 alm (vi) 5x 2 y − 15xy 2 (vii) 10a 2 − 15b 2 c 2 (viii) −4a 2 4ab − 4 ca (ix) x 2 yz xy 2 z xyz 2 (x) ax 2 y bxy 2 cxyz

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A If (x 2 − 4) divides P(x) = 5x 3 kx 2 − x − 36 exactly, fully factorise P(x) and hence obtain the value of k b If x = a is a solution to the equation ax 2 − 5ax 4(2a − 1) = 0, find possible values for a c The polynomials P(x) = x 3 ax 2 bx − 3 and Q(x) = x 3 bx 2 3ax − 9 have a common factor of (x a) Calculate a and b and fully factorise each polynomialFor pq=1 it should have value 1 (checked with some trial values), suggesting that some power of (pq) is a factor Is it possible to factorise it in these terms?→ 3130 ( talk ) 2211, January 14 (UTC)1 h = 6q – 2u (a) Work out the value of h when q = 3 and u = –5 h = (2) B = 3m 2p (b) Work out the value of p when B = 2 and m = 5 p = (2) exemplar (Total for Question 1 is 4 marks) DRAFT D e RAFT xemplar For more interactive questions on this topic visit

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Theresa swims from P to Q, then from Q to R and then finally returns from R to P PQ = 140m, RP = 2m and angle PRQ = 31° (a) Angle PQR is obtuse Calculate its size, to the nearest degree Answer (a) 4 (b) The bearing of Q from P is 060° Calculate the bearing of R from QClick here👆to get an answer to your question ️ Factorize x^3 13x^2 32xKCSE Mathematics Paper 1 Revision Exercise Set 3 Factorise completely the expression #45a^2 – b^2# and hence or otherwise find its value when a= 5 and b= 3

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(i) p 2 6 p 8 (ii) q 2 – 10 q 21 (iii) p 2 6 p – 16 143 Division of Algebraic Expressions We have learnt how to add and subtract algebraic expressions We also12 p is inversely proportional to the square of (q 4) p = 2 when q = 2 Find the value of p when q = –2 Answer p = 3 _____ 13 A car travels a distance of 1280 metres at an average speed of 64 kilometres per hourGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!

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2x =2 × x 4 = 2 × 2= P 2 d Q R (Q2 PR)Vdz is a binomial unit > 1 and this is impossible This completes the proofAnswer (1 of 9) Questions involving nonunity coefficients of p and q (or x, or x and y) require a little bit of trial and error and separate working to generate

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 Another approach is to complete the square, but it is a bit tedious, and the numbers may get too large for rapid mental calculation We need the leading coeffient to be a square, and the coefficient of the n term to be even So we have to multiply this expression by 12 Thus 36 n 2 − 12 ⋅ 53 n 12 ⋅ 8 ⋅ 29Question 1 Factorise the following expressions (i) a 2 8a 16 (ii) p 2 − 10p 25 (iii) 25m 2 30m 9 (iv) 49y 2 84yz 36z 2 (v) 4x 2 − 8x 4 (vi) 121b 2 − bc 16c 2 (vii) (l m) 2 − 4lm (Hint Expand (l m) 2 first)(viii) a 4 2a 2 b 2 b 4 Answer (i) a 2 8a 16 = (a) 2 2 × a × 4 (4) 2 = (a 4) 2 (x y) 2 = x 2 2xy y 2(ii) p 2 − 10p 25 = (p) 2It is not obvious what their factors are We need to develop systematic methods to factorise these expressions, ie, to find their factors This is what we shall do now 1421 Method of common factors • We begin with a simple example Factorise 2 x 4 We shall write each term as a product of irreducible factors;

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