Question

Using factor theorem, factorize each of the following polynomials

$2x^4-7x^3-13x^2$ + 63x - 45

Answer 1

Let f(x) = $2x^4-7x^3-13x^2$ + 63x - 45

Factors of constant term -45 are

$\pm 1,\pm 3,\pm 5,\pm 9,\pm 15,and\pm 45$

Let x = 1, then

f(1) $=2(1{)}^4-7(1{)}^3-13(1{)}^2$ + 63(1)-45

= $2\times 1-7\times 1-13\times 1+63\times$ 1-45

= 2 - 7 - 13 + 63-45=65-65=0

$\therefore$ x - 1 is a factor of f(x)

Let x =3, then

f(3) $=2(3{)}^4-7(3{)}^3-13(3{)}^2$ + 63(3)-45 =162-189-117+189-45 =351-351=0

$\therefore$ x - 3 is its factor

Let x = 5, then

f(5) $=2(5{)}^4-7(5{)}^3-13(5{)}^2$ + 63(5)-45

=1250-875-325+315-45

=1565-1245=320$\ne$ 0

$\therefore$ x -5 is not its factor

Let x = -3, then

f(-3) $=2(-3{)}^4-7(-3{)}^3-13(-3{)}^2+63\times$(1)-45

=162+189-117-189-45

=351-351=0

$\therefore$ x + 3 is a factor of f(x)

Now dividing f(x) by (x-1) (x-3) (x+3)

i.e., dividing $2x^4-7x^3-13x^2+63x-45$ by

$(x-1)(x^2-9)or{x}^3-x^2-9x+9$ we get

Hence, f(x) = (x-1) (x-3) (x+3)(2x-5)