Which of these polynomials could have (x-2) as a factor?

A. $ a(x) = 6x^2 - 7x - 5 $

B. $ b(x) = 3x^2 + 15x - 42 $

C. $ c(x) = 2x^3 + 13x^2 + 16x + 5 $

D. $ d(x) = 3x^3 - 2x^2 - 15x + 14 $

E. $ e(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70 $

F. $ f(x) = x^4 + 5x^3 - 27x^2 - 101x - 70 $

Question

High School

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Answer :

absor201 absor201 · Answer 1 · 2024-02-07T14:12:24-05:00

Answer:

x-2 is a factor of following polynomials:

Step-by-step explanation:

In order to check that which polynomials might have x-2 as a factor, we will put

x-2 = 0 => x = 2 in each polynomial, if the value of polynomial on 2 is zero then x-2 is a factor otherwise not.

So,

a(x) = 6x² - 7x - 5

Putting x = 2

[tex]a(2) = 6(2)^2-7(2)-5\\= 6(4)-14-5\\=24-14-5\\=24-19\\=5 \neq 0[/tex]

b(x) = 3x² + 15x - 42

[tex]b(2) = 3(2)^2 + 15(2) - 42\\=3(4)+30-42\\=12+30-42\\=42-42 = 0[/tex]

c(x) = 2x³ + 13x² + 16x + 5

[tex]c(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5\\=2(8)+13(4)+32+5\\=16+52+32+5\\=105 \neq 0[/tex]

d(x) = 3x³ - 2x² - 15x + 14

[tex]d(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14\\= 3(8)-2(4)-30+14\\=24-8-30+14\\=38-38 = 0[/tex]

e(x) = 8x⁴ - 41x³ - 18x² + 101x + 70

[tex]e(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70\\= 8(16)-41(8)-18(4)+202+70\\=128-328-72+202+70\\=0[/tex]

[tex]f(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70\\=16+5(8)-27(4)-202-70\\=16+40-108-202-70\\=-324 \neq 0[/tex]

Hence,

x-2 is a factor of following polynomials: