There we discussed about Remainder Theorem, Factor Theorem and explained the Method of factoring with Example.
Here, we will apply the method to solve problems. The knowledge of Synthetic Division is made use of here. That knowledge is a prerequisite here. So, please learn the method before proceeding further.
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Let f(x) = x4 + 5x3 + 5x2 - 5x - 6 Here constant term = -6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check witha = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not. f(1) = 1 + 5 + 5 - 5 - 6 = 0 ⇒ x -(1) = (x - 1) is a factor of f(x). Now let us divide f(x) = (x4 + 5x3 + 5x2 - 5x - 6 ) by (x - 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)
Let p(x) = x3 + 6x2 + 11x + 6This is a third Degree Polynomial, which is to be factorized using Facor Theorem. Here constant term = 6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check witha = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not. p(1) is obviously not zero, as all the coefficients are positive. p(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6 = -1 + 6 -11 +6 = 0 ⇒ x -(-1) = (x + 1) is a factor of p(x). Now let us divide p(x) = (x3 + 6x2 + 11x + 6) by (x + 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)
∴ Quotient = (+1)x2 + (+5)x + (6) = x2 + 5x + 6 Thus, p(x) ÷ (x + 1) = x2 + 5x + 6 ∴ p(x) = (x + 1)(x2 + 5x + 6) We can factorize x2 + 5x + 6 using the knowledge of factoring Polynomials of second degree. x2 + 5x + 6 = x2 + 2x + 3x + 6= x(x + 2) + 3(x + 2) = (x + 2)(x + 3) ∴ p(x) = (x + 1)(x + 2)(x + 3) ∴ f(x) = (x - 1)p(x) = (x - 1)(x + 1)(x + 2)(x + 3). Thus, x4 + 5x3 + 5x2 - 5x - 6 = (x - 1)(x + 1)(x + 2)(x + 3). Thus Factoring Polynomial x4 + 5x3 + 5x2 - 5x - 6 by using Factor Theorem gave the Factors as(x - 1)(x + 1)(x + 2)(x + 3). Ans. Thus, the problem on Math Factoring is solved.
Example 2 of Math Factoring
Solve the following problem on Math Factoring
Factorize 4x4 - 12x3 + 7x2 + 3x - 2
Solution to Example 2 of Math Factoring :
Let f(x) = 4x4 - 12x3 + 7x2 + 3x - 2 Here constant term = -2. Its factors are +1,-1,+2, -2. So we check witha = +1,-1,+2, -2, whether f(a) is zero or not. f(1) = 4 - 12 + 7 + 3 - 2 = 0 < BR> ⇒ x - (1) = (x - 1) is a factor of f(x). Now let us divide f(x) = (4x4 - 12x3 + 7x2 + 3x - 2) by (x - 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)
Let p(x) = 4x3 - 8x2 - x + 2This is a third Degree Polynomial, which is to be factorized using Facor Theorem. Here constant term = 2. Its factors are +1,-1,+2, -2. So we check witha = +1,-1,+2, -2, whether f(a) is zero or not. p(1) = 4 - 8 -1 + 2 = -3 ≠ 0 p(-1) = 4(-1)3 - 8(-1)2 -(-1) + 2 = -4 - 8 + 1 + 2 ≠ 0 p(2) = 4(2)3 - 8(2)2 -(2) + 2 = 32 - 32 - 2 + 2 = 0. ⇒ x -(2) = (x - 2) is a factor of p(x). Now let us divide p(x) = (4x3 - 8x2 - x + 2) by (x - 2) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)
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Do the Math Factoring of x4 + 3x3 - 7x2 - 27x - 18
For Answers of these problems on Math Factoring, See at the bottom of the Page.
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