Factorisation Quadratic Polynomial

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 as the difference of two squares.

Question Factorize ^x2– 31x + 220.

Solution: x²– 31a + 220

Question Factorize: –10x² + 31x – 24

Solution:

Question 81a²b²c²+ 64a⁶b²– 144a⁴b²c

Solution: 81a²b²bc² + 64a⁶b²– 144a⁴b²c

= [9abc] ²– 2 [9abc][8a³b] + [8a³b]²

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

Question

(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.

Question Factorize: .

Solution: Here

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

∴

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

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

∴ we have   p(-3)= 0

⇒

Ex. Factorize: 2x² + 12 + 35.

Sol. 2x² + 12 + 35

Product ac = 70 & b = 12

 .

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:

Solution:

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:

Solution:

Ans.

Question Factorize:

Solution: The given expression is

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

Clearly, 6 + 3 = 9 and 6 × 3 = 18.

∴

Question Factorize: by splitting the middle term:

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

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

(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 :

Solution:

Question Factorize:

Solution:

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 .

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) = 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 and the value of c are limited to the factors of 6, which are Hence, the possible rational roots of f(x) are 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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