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Difficulty: MediumFactoring Polynomials

The cubic polynomial 2x3+5x2−8x−202x^3 + 5x^2 - 8x - 20 can be factored completely over the integers into the form (x−a)(x+b)(cx+d)(x - a)(x + b)(cx + d), where aa, bb, cc, and dd are positive integers. What is the value of a+b+c+da + b + c + d?

Answer: 11

Answer

The value of a+b+c+da + b + c + d is 11.
The polynomial 2x3+5x2−8x−202x^3 + 5x^2 - 8x - 20 can be factored completely by first grouping the terms as x2(2x+5)−4(2x+5)=(x2−4)(2x+5)x^2(2x + 5) - 4(2x + 5) = (x^2 - 4)(2x + 5). Factoring the difference of squares x2−4x^2 - 4 yields (x−2)(x+2)(2x+5)(x - 2)(x + 2)(2x + 5). Matching this to the given form (x−a)(x+b)(cx+d)(x - a)(x + b)(cx + d) gives the positive integers a=2a = 2, b=2b = 2, c=2c = 2, and d=5d = 5. The sum of these values is 2+2+2+5=112 + 2 + 2 + 5 = 11.

Step-by-Step Solution

1
Group the terms of the polynomial 2x3+5x2−8x−202x^3 + 5x^2 - 8x - 20.
(2x3+5x2)−(8x+20)(2x^3 + 5x^2) - (8x + 20)
Grouping the terms allows us to look for common factors within each pair of terms.
2
Factor out the greatest common factor (GCF) from each group.
x2(2x+5)−4(2x+5)x^2(2x + 5) - 4(2x + 5)
The GCF of the first group 2x3+5x22x^3 + 5x^2 is x2x^2, and the GCF of the second group 8x+208x + 20 is 44.
3
Factor out the common binomial factor (2x+5)(2x + 5).
(x2−4)(2x+5)(x^2 - 4)(2x + 5)
Both terms share the binomial factor (2x+5)(2x + 5).
4
Factor the difference of squares x2−4x^2 - 4.
(x−2)(x+2)(2x+5)(x - 2)(x + 2)(2x + 5)
The term x2−4x^2 - 4 is a difference of squares, which factors as (x−2)(x+2)(x - 2)(x + 2).
5
Compare the factored expression with the template (x−a)(x+b)(cx+d)(x - a)(x + b)(cx + d) to determine the values of aa, bb, cc, and dd.
a=2a = 2, b=2b = 2, c=2c = 2, and d=5d = 5
Comparing the terms yields x−a=x−2  ⟹  a=2x - a = x - 2 \implies a = 2, x+b=x+2  ⟹  b=2x + b = x + 2 \implies b = 2, and cx+d=2x+5  ⟹  c=2,d=5cx + d = 2x + 5 \implies c = 2, d = 5. All values are positive integers as required.
6
Calculate the sum a+b+c+da + b + c + d.
11
Substituting the values of the variables into the expression gives 2+2+2+5=112 + 2 + 2 + 5 = 11.

Key Concept

Factoring a cubic polynomial by grouping and then factoring the resulting difference of squares.

Alternative Method

Instead of factoring by grouping, we can use the Rational Root Theorem to find rational roots of the polynomial. The possible rational roots of 2x3+5x2−8x−20=02x^3 + 5x^2 - 8x - 20 = 0 are of the form ±pq\pm \frac{p}{q}, where pp is a factor of 2020 and qq is a factor of 22. Testing values shows that x=2x = 2 and x=−2x = -2 are roots, which corresponds to the linear factors (x−2)(x - 2) and (x+2)(x + 2). Dividing the original cubic by their product, (x2−4)(x^2 - 4), yields the remaining linear factor (2x+5)(2x + 5).
Estimated Time:1m 30s
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Topics

Factoring Polynomials

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