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.

Example 1 of Algebra 2 Factoring

Solve the following problem on Algebra 2 Factoring

Factorize 2x^{3} - 7x^{2} - 10x + 24

Solution to Example 1 of Algebra 2 Factoring

Let f(x) = 2x^{3} - 7x^{2} - 10x + 24 Here constant term = +24. Its factors are +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, +8, -8, +12, -12, +24, -24. So we check witha = +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, +8, -8, +12, -12, +24, -24 whether f(a) is zero or not. f(1) = 2 - 7 -10 +24 = +9 ≠ 0; f(-1) = 2(-1)^{3} - 7(-1)^{2} -10(-1) -12 = -2 - 7 + 10 -12 = -11 ≠ 0; f(2) = 2(2)^{3} - 7(2)^{2} - 10(2) + 24 ≠ 0; f(-2) = 2(-2)^{3} - 7(-2)^{2} - 10(-2) + 24 = -16 -28 +20 + 24 = 0; ⇒ x -(-2) = (x + 2) is a factor of f(x). Now let us divide f(x) = (2x^{3} - 7x^{2} - 10x + 24) by (x + 2) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

∴ (2x^{3} - 7x^{2} - 10x + 24) = (x + 2)(x - 4)(2x - 3). Thus Factoring Polynomial 2x^{3} - 7x^{2} - 10x + 24 by using Factor Theorem gave the Factors as(x + 2)(x - 4)(2x - 3). Ans.

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Solve the following problem on Algebra 2 Factoring

Find a and b in order that x^{3} - 6x^{2} + ax + b is exactly divisible by x^{2} - 3x + 2

Solution to Example 3 of Algebra 2 Factoring

Let f(x) = x^{3} - 6x^{2} + ax + bWe can factorize x^{2} - 3x + 2 using the knowledge of factorization of Quadratic Polynomials. x^{2} - 3x + 2 = x^{2} - x - 2x + 2= x(x - 1) - 2(x - 1) = (x - 1)(x - 2) f(x)[ = x^{3} - 6x^{2} + ax + b] is exactly divisible by x^{2} - 3x + 2 ⇒ f(x) is exactly divisible by (x - 1)(x - 2) ⇒ (x - 1) and (x - 2) are factors of f(x) ⇒ f(1) = 0 and f(2) = 0 f(1) = 1 - 6 + a + b = a + b - 5 = 0 ⇒ a + b = 5...........(i) f(2) = (2)^{3} - 6(2)^{2} + a(2) + b = 8 - 24 + 2a + b = 2a + b - 16 f(2) = 0 ⇒ 2a + b - 16 = 0 ⇒ 2a + b = 16..........(ii) (ii) - (i) gives (2a + b) - (a + b) = 16 - 5 ⇒ 2a + b - a - b = 11⇒ a = 11 Using this value of a in (i), we get 11 + b = 5 ⇒ b = 5 - 11 = -6 Thus a = 11, b = -6. Thus Factor Theorem in Factoring Polynomials helped us to find a and b,for x^{3} - 6x^{2} + ax + b to be exactly divisible by x^{2} - 3x + 2 as a = 11 and b = -6. Ans.

Exercise on Algebra 2 Factoring

Solve the Following Problems on Algebra 2 Factoring :

Factorize x^{3} + 6x^{2} + 11x + 6

Factorize 2x^{3} - x^{2} - 15x + 18

Find l and m in order that x^{4} - x^{3} + lx^{2} + mx + 4 is exactly divisible by x^{2} - x - 2

For Answers See at the bottom of the Page.

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