= P 2 d Q R (Q2 PR)Vdz is a binomial unit > 1 and this is impossible This completes the proof7 4(p – q) (4 p 3q) = 16 p2 – 4pq – 12q 2 8 7(x 3 y) (5 x – 2y) = 35 x2 91 xy – 42 y2 Factorisation Worksheet 1 Take out the common factors from each of the following 1 qt rt = t(q r) 2 rt – 5rs = r(t – 5s) 3 3xy 4 xz = x(3 y 4z) Using difference of two squares factorise the following 4 p2 q2 = (p q)(p (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 −
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Factorise (p+q)^2-20(p+q)-125-Factorising quadratic expression of the forms px 2 q where p and q are perfect square Use the identity a 2 b 2 = ( a b )( a b ) , the difference between two squares Example$\endgroup$ – Mark Bennet Oct 2 ' at 1139
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Mathematics NCERT Grade 8, Chapter 14 Factorisation The chapter lays emphasis on the concept of how to express algebraic expressions as the products of their factors In the introduction part following topics are discussed 1 Factors of natural numbers 2 Factors of algebraic expressionsThe Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*xp Note that q and p originate from P/Q reduced to its lowest terms In our case this means that 3x 3x 2x12 can be divided by 3 different polynomials,including by x3 Polynomial Long Division 24 Polynomial Long Division Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry Introduction to Trigonometry Class 10 Important Questions Very Short Answer (1 Mark) Question 1 If tan θ cot θ = 5, find the value of tan2θ cotθ (12)
P = 4 Supplement Solving Quadratic Equation Directly Solving p 2 p = 0 directly Earlier we factored this polynomial by splitting the middle term let us now solve the equation by Completing The Square and by using the Quadratic Formula Parabola, Finding the Vertex 31 Find the Vertex of y = p 2 p3 1 –1 p 24 3 6 18 3 p 1 2 6 p 42 3 p Since 3 x = is a root, the remainder is zero 42 3 0 3 42 14 p p p = = = Note This is just the same synthetic division procedure we are used to Higher Mathematics Unit 2 – Polynomials and Quadratics Page HSN0 hsn uknet 2Solution Steps { x }^ { 3 } 3 { x }^ { 2 } 4 = 0 x 3 − 3 x 2 4 = 0 By Rational Root Theorem, all rational roots of a polynomial are in the form \frac {p} {q}, where p divides the constant term 4 and q divides the leading coefficient 1 List all candidates \frac {p} {q} By Rational Root Theorem, all rational roots of a polynomial are
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^{210 28 Solve the following equations 4 a) 5 x5− 2x− 1 x2−25= 15 16 b) 2x1 2x1−3 2x−1 6−2x2= 11 26 29 Cost of an adult ticket for the movie is $ 16 and one children ticket costs half of the adult's ticket price a) Calculate the total cost of 197 adult tickets and 22 children tickets Upload your study docs or become aThis 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;
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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) $\begingroup$ Look at $(2X3Y)^2$ $\endgroup$ – Damien Oct 2 ' at 1137 2 $\begingroup$ Can you factorise $4x^212x9$?Click here👆to get an answer to your question ️ Factorize x^3 13x^2 32x
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Topic 3 H Simplifying and Substitution St Paul's Catholic School 1 Section A Substitution Grade F / E 1 (a) Find the value of 3x 4y (i) when x = 2 and y = 5 AnswerFor 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)2x =2× x 4 =2 × 2 Hence 2x 4 = (2 × x) (2 × 2) Notice that factor 2 is common to both the terms
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(b) p = 5 # 109 q = 9 # 1016 Expressing each answer in standard form, find (i) p # q, Answer 1 (ii) q Answer 1 17 (a) Find 110% of 70 Answer 1 (b) When new, a car was worth $15 000 After one year it was worth $12 000(P^2q^2r^2)^2 4p^2q^2 =(p^2q^2r^2)^2 (2pq)^2 =(p^2q^2r^22pq)(p^2q^2r^2–2pq) =(p^2q^22pqr^2)(p^2q^2–2pqr^2) ={(pq)^2 r^2}{(pq)^2 r^2} =(pqThe 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 × y
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Finally, we will see how this can be used to fully factorise a polynomial expression and use the factorised version to sketch the curve Play The final video in the series on polynomials We''ll be looking at factor theorem and how we can use it to help factorise cubic and even quartic (power 4) equations (p=6\) and \(q=14\) \(p=2\) andThe PDF's of this chapter are available here, students can download for free from the links given below Chapter 7 Factorization contains nine exercises and the RD Sharma Solutions present in this page provide solutions for the questions present in each exercise Now, let us have a look at the concepts discussed in this chapter Factorise `(pq)^3(qr)^3(rp)^3` Factorise `(pq)^3(qr)^3(rp)^3` Books Physics NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless Chemistry NCERT P Bahadur IITJEE Previous Year Narendra Awasthi MS Chauhan 2 If ,
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Find the quotient of each of the following a 5mn n 2 d 12 xy 2 4 x 2 yz b 12ab 4ab c 3st ( 12s 2 ) e 3 pqr 9 p 2 q 2 pr d 12 pq 4q 2 p 2 e 10a 2bc 5ab 2c 7Solution Steps 5 { x }^ { 3 } { x }^ { 2 } x4=0 5 x 3 − x 2 − 2 0 x 4 = 0 By Rational Root Theorem, all rational roots of a polynomial are in the form \frac {p} {q}, where p divides the constant term 4 and q divides the leading coefficient 5 List all candidates \frac {p} {q} By Rational Root Theorem, all rational roots of a Factorize p²q – pr² – pq r² Solution Given p²q – pr² – pq r 2 Grouping similar terms together we get = p 2 q – pq – pr 2 r 2 Taking the similar terms common we get = pq (p – 1)r 2 (p – 1) Therefore, as (p – 1) is common = (p – 1) (pq –
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Perfect Square Trinomials We may check the pattern of the expression to determine whether it is a Perfect Square Trinomial, namely, p 2 2pq q 2 = ( p q )Answer (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 2Thus 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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(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 followingAccording to remainder theorem f (2) = 0 so that (x – 2 ) is a factor of x3 2x2 x 2 Here maximum power of x is 3 so that its can have maximum 3 factors So our answer is (x1) (x1) (x2) (ii) x3 3x2 9x 5 Possible zeros are factors of ± constant term / coefficient of leading term Here constant term is 5 and coefficient of Factorise(pq)^3(qr)^3(rp)^3 on studyassistantincom There are 60 boys in a class one boy weighing 50 kg goes away and at the same time another boy joins the classes the average weight of the boys is then increased by 1/4 kg what
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Answer Stepbystep explanationLet p(x) = x3 13x2 32x p(1) = 1 13 32 = 33 33 = 0 Therefore (x 1) is a factor of p(x)Factorise 25k^2(4/k^2) math factorise 15x 6x5y2 Maths how would you factorise this (pq)^25(pq) to get this answer(pq)(pq5) Maths Factorise into quadratics X^24x21 X^24x3 X^29x36 Maths how do you factorise 2xyysqaured?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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P^3 9q^3(pq) ^3= =2p^3 10q^3–3p^2*q3p*q^2= =(2pb*q)(p^2cq^2dpq)= =2p^32d*p^2*q2cp*q^2bc*q^3bp^2*qbd*p*q^2 =2p^3(2db)*p^2*q(2cbd)p*q^2bc*q^3 So The first step of factorizing an expression is to take out any common factors which the terms have We consider the algebraic expression 4x8 This expression is already simplified but notice that 4 and 8 have a common factor In fact the HCF of 4 and 8 is 4 Now, 8 = 4 x 2 and 4 = 4 x 1 So, 4x 8 = 4 x 1x – 4 x 2 = 4 (x 2) 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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(iii) 14pq, 28p 2 q 2 Factors of 14pq, 28p 2 q 2 ⇒ 14pq = 2 × 7 × p × q ⇒ 28p 2 q 2 = 2 × 2 × 7 × p × p × q × q So, common factors are ⇒ 2 × 7 × p × q = 14pq (iv) 2x, 3x 2, 4 Factors of 2x, 3x 2, 4 ⇒ 2x = 2 × x ⇒ 3x 2 = 3 × x × x ⇒ 4 = 2 × 2 So, common factor is 1 (∵ 1 is a factor of every number) (v) 6abc, 24ab Class 10 Maths MCQs Chapter 5 Arithmetic Progressions 1 The n th term of an AP is given by a n = 3 4n The common difference is 2 If p, q, r and s are in AP then r – q is 3 If the sum of three numbers in an AP is 9 and their product is 24, then numbers are Also (a – d)ML Aggarwal Solutions PDF of this chapter can be downloaded from the link provided below and can be used for any future references Chapter 10 Algebraic Expressions and Identities contains five exercises and Check your progress questions The ML Aggarwal Class 8 Solutions present on this page provides clear solutions, according to the latest
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12 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 hourIn general, it is true that every polynomial splits (ie, factors into linear factors) over the complex number field However, finding this factorization for polynomials of degree up to four is manageable, and factoring something of degree higher than 4 is impossible (in the sense that there does not exist, for a general polynomial of degree 5 or higher, an algorithm for finding its roots) 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
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