7.2 Factor quadratic trinomials with leading coefficient 1  (Page 2/3)

Solution

Again, with the positive last term, 28, and the negative middle term, $\mathrm{-11}t$ , we need two negative factors. Find two numbers that multiply 28 and add to $\mathrm{-11}$ .

$\begin{array}{cccc}& & & t^2-11t+28\hfill \\ \\ \text{Write the factors as two binomials with first terms}t.\hfill & & & \left(t\right)\left(t\right)\hfill \end{array}$

Find two numbers that: multiply to 28 and add to $\mathrm{-11}$ .

$\begin{array}{cccc}\text{Use}\mathrm{-4},\mathrm{-7}\text{as the last terms of the binomials.}\hfill & & & \hfill \left(t-4\right)\left(t-7\right)\hfill \\ \text{Check.}\hfill & & & \\ \\ \\ (t-4)(t-7)\hfill & & & \\ t^2-7t-4t+28\hfill & & & \\ t^2-11t+28✓\hfill & & & \end{array}$

Solution

To get a negative last term, multiply one positive and one negative. We need factors of $\mathrm{-5}$ that add to positive 4.

Notice: We listed both $1,\mathrm{-5}\text{and}-1,5$ to make sure we got the sign of the middle term correct.

$\begin{array}{cccc}& & & z^2+4z-5\hfill \\ \text{Factors will be two binomials with first terms}z.\hfill & & & \left(z\right)\left(z\right)\hfill \\ \text{Use}\mathrm{-1},5\text{as the last terms of the binomials.}\hfill & & & \left(z-1\right)\left(z+5\right)\hfill \\ \text{Check.}\hfill & & & \\ \\ \\ \left(z-1\right)\left(z+5\right)\hfill & & & \\ z^2+5z-1z-5\hfill & & & \\ z^2+4z-5✓\hfill & & & \end{array}$

Solution

This time, we need factors of $\mathrm{-5}$ that add to $\mathrm{-4}$ .

$\begin{array}{cccc}& & & z^2-4z-5\hfill \\ \text{Factors will be two binomials with first terms}z.\hfill & & & \left(z\right)\left(z\right)\hfill \\ \text{Use}1,\mathrm{-5}\text{as the last terms of the binomials.}\hfill & & & \left(z+1\right)\left(z-5\right)\hfill \\ \text{Check.}\hfill & & & \\ \\ \\ \left(z+1\right)\left(z-5\right)\hfill & & & \\ z^2-5z+1z-5\hfill & & & \\ z^2-4z-5✓\hfill & & & \end{array}$

Notice that the factors of $z^2-4z-5$ are very similar to the factors of $z^2+4z-5$ . It is very important to make sure you choose the factor pair that results in the correct sign of the middle term.

Solution

$\begin{array}{cccc}& & & q^2-2q-15\hfill \\ \text{Factors will be two binomials with first terms}q.\hfill & & & \left(q\right)\left(q\right)\hfill \\ \text{You can use}3,\mathrm{-5}\text{as the last terms of the}\hfill & & & \left(q+3\right)\left(q-5\right)\hfill \\ \text{binomials.}\hfill & & & \end{array}$

Check.

$\left(q+3\right)\left(q-5\right)$

$q^2-5q+3q-15$

$q^2-2q-15✓$