Chapter 5, Problem 24CT

To determine

To factor: The given polynomial $z^2+z-3$ . If the polynomial is not factorable, write it prime.

Answer to Problem 24CT

The factored polynomial is $z^2+z-3=\frac{1}{2}\cdot \frac{1}{2}\left(2z+1+\sqrt{7}\right)\left(2z+1-\sqrt{7}\right)$ .

Explanation of Solution

Given information:

The given polynomial is $z^2+z-3$ .

Concept used:

The factorization of a polynomials involves breaking up a polynomial into simpler terms (the factors) such that when the terms are multiplied together they equal the original polynomial.

Calculations:

The given polynomial (say P ) is $P\left(z\right)=z^2+z-3$ .

Now, determine factors of -3 whose sum is 1 that is not possible as $-3=-3\times 1=3\times -1$ .

So we try perfect square method and difference of two square. For, $P\left(z\right)=z^2+z-3$ may be written as

  $P\left(z\right)=z^2+z-3=z^2+2\cdot z\cdot \frac{1}{2}+\frac{1}{4}-\frac{1}{4}-3$

  $P\left(z\right)=\left(z^2+2\cdot z\cdot \frac{1}{2}+\frac{1}{4}\right)-\left(\frac{1}{4}+3\right)$

  $P\left(z\right)={\left(z+\frac{1}{2}\right)}^2-\left(\frac{7}{4}\right)$

  $P\left(z\right)={\left(z+\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{7}}{2}\right)}^2$

  $P\left(z\right)=\left(z+\frac{1}{2}+\frac{\sqrt{7}}{2}\right)\left(z+\frac{1}{2}-\frac{\sqrt{7}}{2}\right)$$\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$

  $P\left(z\right)=\frac{1}{2}\cdot \frac{1}{2}\left(2z+1+\sqrt{7}\right)\left(\left(2z+1+\sqrt{7}\right)\right)$

Thus, the factored polynomial is $P\left(z\right)=\frac{1}{2}\cdot \frac{1}{2}\left(2z+1+\sqrt{7}\right)\left(2z+1-\sqrt{7}\right)$ , i.e., $z^2+z-3=\frac{1}{2}\cdot \frac{1}{2}\left(2z+1+\sqrt{7}\right)\left(2z+1-\sqrt{7}\right)$

Conclusion:

The factored polynomial is $z^2+z-3=\frac{1}{2}\cdot \frac{1}{2}\left(2z+1+\sqrt{7}\right)\left(2z+1-\sqrt{7}\right)$ .