Facorisation of following polynomial is
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Answer:
where is the term???
Step-by-step explanation:
Factoring Quadratic Polynomials
Case 1: When the coefficient of x2 is unity.
The general form of an equation is x2+bx+c. Every quadratic equation can be expressed as: x2+bx+c= (x+d)(x+e). Here, b is the sum of d and e & c is the product of d and e. Example: (x+2)(x+3) = x2+ 2x + 3x + 6 = x2+ 5x + 6.
Here, 5 = 2 + 3 = d + e = b in general form and 6 = 2 × 3 = d × e = c in general form. To factorize quadratic polynomial, we shall be looking for numbers which on multiplication will get equal to c and on summation equal to b.
Example: Factorize x2+8x+12.
Solution: Below the steps are given for your understanding.
Step 1: Factorize 12 as 12 = 2 × 6 or = 4 × 3. We have to find a pair, such that its product is equal to 12 and summation is equal to 8. Only one such pair is possible i.e. 2 and 6.
2 + 6 = 8
2 × 6 = 12
Step 2: Break middle term in terms of the summation of a pair of numbers such that its product is equal to c i.e. 12 in above case. We will write 8=6+2
x2+ (6+2)x+ 12
= x2+ 6x +2x + 12
Step 3: Form pairs of terms and factor out GCD of the two pairs separately.
= x2+ 6x +2x + 12 = (x2+ 6x) +(2x + 12) = x(x+6)+2(x+6)
Step 4: Again factor out GCD of remaining sum of products. Follow factorization procedure of binomials as explained earlier. Factor out (x+6) from sum of product,
=x(x+6)+2(x+6) = (x+6)(x+2)
Example: Do the factorization of polynomials: x2-5x-6
Soluiton: Factors of -6 are 2 × -3; – 3 × 2; 1 × -6; -1 × 6 and the pair summing up to -5 is (-6,1) as required. Hence,
x2-5x-6 = x2-6x+x-6 = x(x-6)+(x-6) = (x-6)(x+1)
Case 2:When the coefficient of x2 a is an integer other than 1 or -1
General form is given by ax2+bx+c. Any quadratic of form ax2+bx+c is expressible in the product of two linear polynomials:
ax2+bx+c= (a1x+b1)(a2x+b2)
where a is the product of a1 and a2, c is the product of b1 and b2 and b is the sum of the product of a1b2 and a2b1. Consider one more example,
(3x+2)(x+4)= 3x2+ 2x +12x +8 = 3x2+14x+ 8
Here, 3 = 3 × 1 = a1 × a2 = a in general form. 14 = 2 × 1 + 4 × 3 = (a1b2) × (a2b1) = b in general form. 8 = 4 × 2 = b1 × b2= c in general form. Example: Factorize 3x2 – 5x – 2
Step 1: find factors of 3 × – 2 = -6.
Step 2: Product of factors of -6 are = -3 × 2; -2 × 3; 6 × -1; -1 × 6. Here, the possible pair of factor gives a summation of -1 is (-6,1).
Step 3: Break the middle term as the summation of two numbers such that its product is equal to -6. Calculated above such two numbers are -6 and 1.
Step 4: Breaking the middle term: x2-6x+x-2
Step 5: Making pairs of terms: (3x2-6x)+(x-2)
Step 6: Factor out GCD for both the pairs: 3x(x-2)+1(x-2). Note: 1 is factored out from the second bracket as there was no other common factor.
Step 7: Factor out GCD from the resulting summation of products: =3x(x-2)+1(x-2) =(x-2)(3x+1)