Chapter T, Problem 4ADT

Factor each expression.

1. (a) 4x² − 25
2. (b) 2x² + 5x − 12
3. (c) x³ − 3x² − 4x + 12
4. (d) x⁴ + 27x
5. (e) 3x^3/2 − 9x^1/2 + 6x^−1/2
6. (f) x³y − 4xy

To determine

To factor: The expression $4x^2-25$.

Answer to Problem 4ADT

The expression $4x^2-25$ is factorized as $\left(2x+5\right)\left(2x-5\right)$.

Explanation of Solution

Consider the expression $4x^2-25$.

Simplify the above expression as follows,

$\begin{array}{c}4x^2-25={\left(2x\right)}^2-{\left(5\right)}^2\\ =\left(2x+5\right)\left(2x-5\right)\left(\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right)\end{array}$

Thus, the expression $4x^2-25$ is factorized as $\left(2x+5\right)\left(2x-5\right)$.

To determine

To factor: The expression $2x^2+5x-12$.

Answer to Problem 4ADT

The expression $2x^2+5x-12$ is factorized as $\left(x+4\right)\left(2x-3\right)$.

Explanation of Solution

Consider the expression $2x^2+5x-12$.

Simplify the above expression as follows,

$\begin{array}{c}2x^2+5x-12=2x^2+8x-3x-12\\ =2x\left(x+4\right)-3\left(x+4\right)\\ =\left(x+4\right)\left(2x-3\right)\end{array}$

Thus, the expression $2x^2+5x-12$ is factorized as $\left(x+4\right)\left(2x-3\right)$.

To determine

To factor: The expression $x^3-3x^2-4x+12$.

Answer to Problem 4ADT

The expression $x^3-3x^2-4x+12$ is factorized as $\left(x-3\right)\left(x+2\right)\left(x-2\right)$.

Explanation of Solution

Consider the expression $x^3-3x^2-4x+12$.

Simplify the above expression as follows,

$\begin{array}{c}{x}^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\\ =\left(x-3\right)\left(x^2-4\right)\\ =\left(x-3\right)\left(x+2\right)\left(x-2\right)\left(\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right)\end{array}$

Thus, the expression $x^3-3x^2-4x+12$ is factorized as $\left(x-3\right)\left(x+2\right)\left(x-2\right)$.

To determine

To factor: The expression $x^4+27x$.

Answer to Problem 4ADT

The expression $x^4+27x$ is factorized as $x\left(x+3\right)\left(x^2-3x+9\right)$.

Explanation of Solution

Consider the expression $x^4+27x$.

Simplify the above expression as follows,

$\begin{array}{c}{x}^4+27x=x\left(x^3+27\right)\\ =x\left[{\left(x\right)}^3+{\left(3\right)}^3\right]\\ =x\left(x+3\right)\left(x^2-x\times 3+3^2\right)\left(\because a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)\right)\\ =x\left(x+3\right)\left(x^2-3x+9\right)\end{array}$

Thus, the expression $x^4+27x$ is factorized as $x\left(x+3\right)\left(x^2-3x+9\right)$.

To determine

To factor: The expression $3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}$.

Answer to Problem 4ADT

The expression $3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}$ is factorized as $3x^{-\frac{1}{2}}\left(x-2\right)\left(x-1\right)$.

Explanation of Solution

Consider the expression $3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}$.

Simplify the above expression as follows,

$\begin{array}{c}3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}=3x^{-\frac{1}{2}}\left(x^2-3x+2\right)\\ =3x^{-\frac{1}{2}}\left(x^2-2x-x+2\right)\\ =3x^{-\frac{1}{2}}\left(x\left(x-2\right)-1\left(x-2\right)\right)\\ =3x^{-\frac{1}{2}}\left(x-2\right)\left(x-1\right)\end{array}$

Thus, the expression $3x^{\frac{3}{2}}-9x^{\frac{1}{2}}+6x^{-\frac{1}{2}}$ is factorized as $3x^{-\frac{1}{2}}\left(x-2\right)\left(x-1\right)$.

To determine

To factor: The expression $x^{3}y-4xy$.

Answer to Problem 4ADT

The expression $x^{3}y-4xy$ is factorized as $xy\left(x+2\right)\left(x-2\right)$.

Explanation of Solution

Consider the expression $x^{3}y-4xy$.

Simplify the above expression as follows,

$\begin{array}{c}{x}^{3}y-4xy=xy\left(x^2-4\right)\\ =xy\left[{\left(x\right)}^2-{\left(2\right)}^2\right]\\ =xy\left(x+2\right)\left(x-2\right)\left(\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right)\end{array}$

Thus, the expression $x^{3}y-4xy$ is factorized as $xy\left(x+2\right)\left(x-2\right)$.