Polynomials with 4 Terms

9) [tex]f(x)=x^3+2 x^2+5 x+10[/tex]
A) [tex]f(x)=(x+2)(x^2+5)[/tex]
B) [tex]f(x)=(x+2)(x^2+7)[/tex]
C) [tex]f(x)=(x+1)(x^2+5)[/tex]
D) [tex]f(x)=(x+2)(2 x^2+5)[/tex]
E) [tex]f(x)=x(x^2+5)[/tex]

10) [tex]f(x)=x^3-3 x^2+5 x-15[/tex]
A) [tex]f(x)=(x-3)(x^2+4)[/tex]
B) [tex]f(x)=(x+4)(x^2+5)[/tex]
C) [tex]f(x)=(x-3)(x^2+3)[/tex]
D) [tex]f(x)=(x-3)(x^2+6)[/tex]
E) [tex]f(x)=(x-3)(x^2+5)[/tex]

For the first polynomial f ( x ) = x 3 + 2 x 2 + 5 x + 10 , expand each option and find the matching factorization: A) ( x + 2 ) ( x 2 + 5 ) .
For the second polynomial f ( x ) = x 3 − 3 x 2 + 5 x − 15 , expand each option and find the matching factorization: E) ( x − 3 ) ( x 2 + 5 ) .
The correct factorization for the first polynomial is A.
The correct factorization for the second polynomial is E. A an d E ​

Explanation

Problem Analysis We are given two cubic polynomials and asked to find their correct factorizations from a list of options. We will expand each option and compare it to the original polynomial to find the correct match.

Factorizing the first polynomial For the first polynomial, f ( x ) = x 3 + 2 x 2 + 5 x + 10 , we test each factorization:


A) ( x + 2 ) ( x 2 + 5 ) = x ( x 2 + 5 ) + 2 ( x 2 + 5 ) = x 3 + 5 x + 2 x 2 + 10 = x 3 + 2 x 2 + 5 x + 10 . This matches the original polynomial.
B) ( x + 2 ) ( x 2 + 7 ) = x ( x 2 + 7 ) + 2 ( x 2 + 7 ) = x 3 + 7 x + 2 x 2 + 14 = x 3 + 2 x 2 + 7 x + 14 . This does not match.
C) ( x + 1 ) ( x 2 + 5 ) = x ( x 2 + 5 ) + 1 ( x 2 + 5 ) = x 3 + 5 x + x 2 + 5 = x 3 + x 2 + 5 x + 5 . This does not match.
D) ( x + 2 ) ( 2 x 2 + 5 ) = x ( 2 x 2 + 5 ) + 2 ( 2 x 2 + 5 ) = 2 x 3 + 5 x + 4 x 2 + 10 = 2 x 3 + 4 x 2 + 5 x + 10 . This does not match.
E) x ( x 2 + 5 ) = x 3 + 5 x . This does not match.
Therefore, the correct factorization for the first polynomial is A) f ( x ) = ( x + 2 ) ( x 2 + 5 ) .

Factorizing the second polynomial For the second polynomial, f ( x ) = x 3 − 3 x 2 + 5 x − 15 , we test each factorization:

A) ( x − 3 ) ( x 2 + 4 ) = x ( x 2 + 4 ) − 3 ( x 2 + 4 ) = x 3 + 4 x − 3 x 2 − 12 = x 3 − 3 x 2 + 4 x − 12 . This does not match.
B) ( x + 4 ) ( x 2 + 5 ) = x ( x 2 + 5 ) + 4 ( x 2 + 5 ) = x 3 + 5 x + 4 x 2 + 20 = x 3 + 4 x 2 + 5 x + 20 . This does not match.
C) ( x − 3 ) ( x 2 + 3 ) = x ( x 2 + 3 ) − 3 ( x 2 + 3 ) = x 3 + 3 x − 3 x 2 − 9 = x 3 − 3 x 2 + 3 x − 9 . This does not match.
D) ( x − 3 ) ( x 2 + 6 ) = x ( x 2 + 6 ) − 3 ( x 2 + 6 ) = x 3 + 6 x − 3 x 2 − 18 = x 3 − 3 x 2 + 6 x − 18 . This does not match.
E) ( x − 3 ) ( x 2 + 5 ) = x ( x 2 + 5 ) − 3 ( x 2 + 5 ) = x 3 + 5 x − 3 x 2 − 15 = x 3 − 3 x 2 + 5 x − 15 . This matches the original polynomial.
Therefore, the correct factorization for the second polynomial is E) f ( x ) = ( x − 3 ) ( x 2 + 5 ) .

Final Answer The correct factorizations are:

For f ( x ) = x 3 + 2 x 2 + 5 x + 10 , the correct factorization is A) ( x + 2 ) ( x 2 + 5 ) .
For f ( x ) = x 3 − 3 x 2 + 5 x − 15 , the correct factorization is E) ( x − 3 ) ( x 2 + 5 ) .
Examples
Factoring polynomials is a fundamental skill in algebra. It's like breaking down a complex number into its prime factors. In real life, factoring can help engineers design structures by understanding the polynomial equations that describe the forces acting upon them. For example, when designing a bridge, engineers use polynomial equations to model the load distribution and ensure the bridge's stability. Factoring these polynomials helps them identify critical points and potential weaknesses in the design, ensuring safety and efficiency.

Answered by GinnyAnswer | 2025-07-03