Question

Write each of the polynomial listed below as a product of two first degree polynomials:

(i) 2x² + 5x + 3

(ii) x² + 2x − 1

(iii) x² + 3x + 2

(iv) x² − 2

(v) 4x² + 20x + 25

(vi) x² − x − 1

Answer 1

According to the Factor theorem, to find the first degree factors of a polynomial, we need to find those numbers x, which make the given polynomial zero.

(i)

The first degree factors of the polynomial 2x² + 5x + 3 can be obtained by finding out the solutions of the equation 2x² + 5x + 3 = 0.

_{On comparing with the general quadratic equation, ax}² _{+ bx + c = 0, we have:}

_{a = 2, b = 5 and c = 3}

x =

= −1 or

_{So, by the Factor theorem, (x + 1) and are the factors of the given polynomial.}

Now, (x + 1) = x² + +

This is not the original polynomial.

We can write:

⇒ (x + 1) =

⇒

⇒

(ii)

The first degree factors of the polynomial x² + 2x − 1 can be obtained by finding out the solutions of the equation x² + 2x − 1 = 0.

_{On comparing with the general quadratic equation, ax}² _{+ bx + c = 0, we have:}

_{a = 1, b = 2 and c = −1}

x =

= or

_{So, by the Factor theorem,} (x + ) _and (x + ) _{are the factors of the given polynomial.}

⇒ (x + ) (x + ) = x² + 2x − 1

(iii)

The first degree factors of the polynomial x² + 3x + 2 can be obtained by finding out the solutions of the equation x² + 3x + 2 = 0.

_{On comparing with the general quadratic equation, ax}² _{+ bx + c = 0, we have:}

_{a = 1, b = 3 and c = 2}

x =

=

=

=

= −1 or −2

_{So, by the Factor theorem, (x + 1) and (x + 2) are the factors of the given polynomial.}

⇒ (x + 1) (x + 2) = x² + 3x + 2

(iv)

The first degree factors of the polynomial x² − 2 can be obtained by finding out the solutions of the equation x² − 2 = 0.

x² − 2 = 0

⇒ (x − )(x + ) = 0

⇒ x = or −

_{So, by the Factor theorem,} (x + ) _and (x − ) _{are the factors of the given polynomial.}

⇒ (x + )(x − ) = x² − 2

(v)

The first degree factors of the polynomial 4x² + 20x + 25 can be obtained by finding out the solutions of the equation 4x² + 20x + 25 = 0.

_{On comparing with the general quadratic equation, ax}² _{+ bx + c = 0, we have:}

_{a = 4, b = 20 and c = 25}

x =

=

=

= −

_{So, by the Factor theorem, and are the factors of the given polynomial.}

Now,

This is not the original polynomial.

We can write:

(vi)

The first degree factors of the polynomial x² − x − 1 can be obtained by finding out the solutions of the equation x² − x − 1 = 0.

_{On comparing with the general quadratic equation, ax}² _{+ bx + c = 0, we have:}

_{a = 1, b = −1 and c = −1}

x =

=

=

= or

_{So, by the Factor theorem, and are the factors of the given polynomial.}

⇒