Factorisation Quadratic Polynomial

Polynomial of Class 9

FACTORISATION QUADRATIC POLYNOMIAL

For factorisation of a quadratic expression ax2 + bx + a where a ≠ 0, there are two method.

(a) By Method of Completion of Square :

In the form ax2 + bx + c where a ≠ 0, firstly we take ‘a’ common in the whole expression then factorise by converting the expression FACTORISATION QUADRATIC POLYNOMIAL as the difference of two squares.

FACTORISATION QUADRATIC POLYNOMIAL

Question Factorize x2– 31x + 220.

Solution: x2– 31a + 220

 FACTORISATION QUADRATIC POLYNOMIAL

Question Factorize: –10x2 + 31x – 24

Solution:

 FACTORISATION QUADRATIC POLYNOMIAL

Question 81a2b2c2+ 64a6b2– 144a4b2c

Solution: 81a2b2bc2 + 64a6b2– 144a4b2c

= [9abc] 2– 2 [9abc][8a3b] + [8a3b]2

= [9abc –8a3b]2 = a2b2 [9c – 8a2]2 Ans.

Question FACTORISATION QUADRATIC POLYNOMIAL

 FACTORISATION QUADRATIC POLYNOMIAL

(b) By Splitting the Middle Term:

In the quadratic expression ax2 + bx + c, where a is the coefficient of x2, b is the coefficient of x and c is the constant term. In the quadratic expression of the form x2 + bx + c, a = 1 is the multiple of x2 and another terms are the same as above.

There are four types of quadratic expression with respect to to the sign change of coefficients of x and constant term :

(i) ax2 + bx + c (ii) ax2– bx + c

(iii) ax2– bx – c (iv) ax2 + bx – c

Splitting middle term of ax2 + bx +c.

In this case, we find integers s and t such that  s + t = b and st = ac.

 FACTORISATION QUADRATIC POLYNOMIAL

Question Factorize: FACTORISATION QUADRATIC POLYNOMIAL.

Solution: Here FACTORISATION QUADRATIC POLYNOMIAL

We try to split 5x into two parts whose sum is 5 and product is = –24

Clearly, 8 + (–3) = 5 and 8 × –3 = –24

 Middle term 5x can be written as 8x - 3x

FACTORISATION QUADRATIC POLYNOMIAL

Question Find the value of q, if  X + 3 is a factor of FACTORISATION QUADRATIC POLYNOMIAL

Solution: Let FACTORISATION QUADRATIC POLYNOMIAL be the given polynomial. According to question (X+3) is a factor of  p(x)

∴ we have   p(-3)= 0

FACTORISATION QUADRATIC POLYNOMIAL

 

Ex. Factorize: 2x2 + 12FACTORISATION QUADRATIC POLYNOMIAL + 35.

Sol. 2x2 + 12FACTORISATION QUADRATIC POLYNOMIAL + 35

Product ac = 70 & b = 12FACTORISATION QUADRATIC POLYNOMIAL

 FACTORISATION QUADRATIC POLYNOMIAL.

Question Factorize: x2 – 14x + 24.

Solution: Product ac = 24 & b = -14

∴ Split the middle term as – 2 &– 2

⇒ x2– 14x + 24 = x2– 12 – 2x + 24

⇒ x(x – 12) – 2 (x – 12) = (x – 12)(x – 2) Ans.

Question Factorize: FACTORISATION QUADRATIC POLYNOMIAL

Solution: FACTORISATION QUADRATIC POLYNOMIAL

Product ac = – 48 & b = – 13 ∴ We split the middle term as – 16x + 3x.

=1/24[24x2– 16x + 3x – 2]

=1/24[8x(3x – 2) + 1(3x – 2)]

=1/24(3x – 2)(8x + 1) Ans.

Question Factorize:FACTORISATION QUADRATIC POLYNOMIAL

Solution: FACTORISATION QUADRATIC POLYNOMIAL

FACTORISATION QUADRATIC POLYNOMIAL Ans.

Question Factorize: FACTORISATION QUADRATIC POLYNOMIAL

Solution: The given expression is FACTORISATION QUADRATIC POLYNOMIAL

We try to split 9 into two parts whose sum is 9 and product 18.

Clearly, 6 + 3 = 9 and 6 × 3 = 18.

FACTORISATION QUADRATIC POLYNOMIAL

Question Factorize: FACTORISATION QUADRATIC POLYNOMIAL by splitting the middle term:

Solution: In order to factorize FACTORISATION QUADRATIC POLYNOMIAL we try to split 3√3 into two parts whose sum is 3√3 and product is 6.

 FACTORISATION QUADRATIC POLYNOMIAL

So, we write the middle term 3√3 as 2√3x + √3x

 FACTORISATION QUADRATIC POLYNOMIAL

(c) Factorisation by Using the Formula for the Difference of Two Squares:

a2– b2 = ( a + b) (a - b)

Ex. Factorise:4(2a + 3b – 4c) 2– (a – 4b + 5c)2

Solution: = 4(2a + 3b –4c)2– (a – 4b + 5c) 2

= [2(2a + 3b – 4c)]2– (a – 4b + 5c)2

= [4a + 6b – 8c + a – 4b + 5c] [4a + 6b – 8c – a + 4b – 5c]

= [5a + 2b – 3c] [3a + 10b – 13c] Ans.

Question Factorise :FACTORISATION QUADRATIC POLYNOMIAL

Solution: FACTORISATION QUADRATIC POLYNOMIAL

FACTORISATION QUADRATIC POLYNOMIAL

Question Factorize: FACTORISATION QUADRATIC POLYNOMIAL

Solution: FACTORISATION QUADRATIC POLYNOMIAL

FACTORISATION QUADRATIC POLYNOMIAL Ans.

Question Factorize: x4 + x2y2 + y4.

Sol. x2 + x2y2 + y4 = (x2) 2 + 2.x2.y2 + (y2)2– x2y2

= x2 + y2)2– (xy)2

= (x2 + y2 + xy) (x2 + y2– xy) Ans.

(d) Factorization by Using Formula of a3 + b3 and a3– b3:

Ex. Factorize: 64a13b + 343ab13.

Solution: 64a13b + 343ab13 = ab[64a12 + 343b12]

= ab[(4a4)3 + (7b4)3]

= ab[4a4 + 7b4] [(4a4)2– (4a4) (7b4) + (7b4)2]

= ab[4a4 + 7b4][16a8– 28a4b8 + 49b8] Ans.

Question Factorize: p3q2x4 + 3p2qx3 + 3px2 + X/q– q2r3x

Solution: In above question, If we take common then it may become in the form of 3 + b3.

 p3q2x4 + 3p2qx3 + 3px2 + X/q– q2r3x

= X/q[p3q3x3 + 3p2q2x2 + 3pqx + 1 – q3r3]

= X/q[(pqx)3 + 3(pqx)2 .1 + 3pqx . (1)2 + (1)3– q3r3]

Let pqx = A & 1 = B

= X/q [A3 + 3A2B + 3AB2 + B3– q3r3]

= X/q[(pqx + 1)3– (qr)3] = X/q[pqx + 1 – qr][(pqx + 1)2 + (pqx + 1) qr + (qr)3]

= X/q[pqx + 1 – qr][p2q2x2 + 1 + 2pqx + pq2xr + qr + q2r2] Ans.

Question Factorize: x3– 6x2 + 32

Solution: x3 + 32 – 6x2

= x3 + 8 + 24 – 6x2

= [(x)3 + (2)3] + 6[4 – x2]

= (x + 2)[x2– 2x + 4] + 6[2 + x][2 – x]

= (x + 2) [x2– 2x + 4 + 6(2 – x)]

= (x + 2)[x2– 2x + 4 + 12 – 6x]

= (x + 2) [x2– 8x + 16]

= (x + 2) (x – 4)2 Ans.

(e) Integral Root Theorem :

If f(x) is a polynomial with integral coefficient and the leading coefficient is 1, then any integer root of f(x) is a factor of the constant term. Thus if f(x) = x3– 6x2 + 11x - 6 has an Integral root, then it is one of the factors of 6 which are FACTORISATION QUADRATIC POLYNOMIAL.

Now Infect f(1) = (1)3– 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 =0

f(2) = (2)3– 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0

f(3) = (3)3– 6(3)2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0

Therefore Integral roots of f(x) are 1,2,3.

(f) Rational Root Theorem :

Let b/c be a rational fraction in lowest terms. If b/c is a root of the polynomial f(x) = FACTORISATION QUADRATIC POLYNOMIAL with integral coefficients. Then b is a factor of constant term a0 and c is a factor of the leading coefficient an.

For example :If b/c is a rational root of the polynomial f(x) = 6x3 + 5x2– 3x – 2, then the values of b are limited to the factors of –2 which are FACTORISATION QUADRATIC POLYNOMIAL and the value of c are limited to the factors of 6, which are FACTORISATION QUADRATIC POLYNOMIAL Hence, the possible rational roots of f(x) are FACTORISATION QUADRATIC POLYNOMIAL Infect -1 is a Integral root and  2/3,-1/2 are the rational roots of f(x) = 6x3 + 5x2– 3x – 2.

NOTE:

(i) An nth degree polynomial can have at most n real roots.

(ii) Finding a zero or root of polynomial f(x) means solving the polynomial equation f(x) = 0. It follows from the above discussion that if f(x) = ax + b,  a ≠ 0 is a linear polynomial, then it has only one root given by f(x) = 0 i.e. f(x) = ax + b = 0

⇒ ax = – b

⇒ x =

Thus   a=  is the only root of f(x) = ax + b.

Question If f(x) = 2x3– 13x2 + 17x + 12 then find out the value of f(-2) & f(3).

Solution: f(x) = 2x3– 13x2 + 17x + 12

f(-2) = 2(–2)3– 13(–2)2 + 17 (–2) + 12

= –16 – 52 – 34 + 12 = – 90 Ans.

f(3) = 2(3)3– 13(3)2 + 17(3) + 12

= 54 – 117 + 51 + 12 = 0 Ans.

(f) Factorisation of an Expression Reducible to a Quadratic Expression :

Question Factorize:- 8 + 9(a – b)6– (a – b)12

Solution: –8 + 9(a –b)6– (a – b)12

Let (a –b)6 = x

Then –8 + 9x – x2 = – (x2– 9x + 8) = – (x2– 8x – x + 8)

= – (x– 8)(x – 1)

= – [(a– b)6– 8][(a – b)6– 1]

= [1 – (a –b)6][(a – b)6– 8]

= [(1)3– {(a – b)2}3][{(a – b)2}3– (2)3]

= [1 – (a –b)2][1 + (a – b)4 + (a – b)2][(a – b)2– 2][(a – b)4 + 4 + 2(a – b)2] Ans.

Question Factorize: 6x2– 5xy – 4y2 + x + 17y – 15

Solution: 6x2 + x[1 – 5y] – [4y2– 17y + 15]

= 6x2 + x[1 – 5y] – [4y2– 17y + 15]

= 6x2 + x[1 – 5y] – [4y(y – 3) – 5(y – 3)]

= 6x2 + x[1 – 5y] – (4y – 5)(y – 3)

= 6x2 + 3(y –3)x– 2(4y – 5)x – (4y –)(y - 3)

= 3x[2x + y 3] - (4y - 5)(2x + y - 3)

= (2x + y - 3)(3x - 4y + 5) Ans.

 

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