Solve -(3t-1)^2-(3t+2)(3t-2)

$$- ( 3 t - 1 ) ^ { 2 } - ( 3 t + 2 ) ( 3 t - 2 )$$

Evaluate

$$3+6t-18t^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-1\right)^{2}.

$$-\left(9t^{2}-6t+1\right)-\left(3t+2\right)\left(3t-2\right)$$

To find the opposite of 9t^{2}-6t+1, find the opposite of each term.

$$-9t^{2}+6t-1-\left(3t+2\right)\left(3t-2\right)$$

Consider \left(3t+2\right)\left(3t-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$-9t^{2}+6t-1-\left(\left(3t\right)^{2}-4\right)$$

Expand \left(3t\right)^{2}.

$$-9t^{2}+6t-1-\left(3^{2}t^{2}-4\right)$$

Calculate 3 to the power of 2 and get 9.

$$-9t^{2}+6t-1-\left(9t^{2}-4\right)$$

To find the opposite of 9t^{2}-4, find the opposite of each term.

$$-9t^{2}+6t-1-9t^{2}+4$$

Combine -9t^{2} and -9t^{2} to get -18t^{2}.

$$-18t^{2}+6t-1+4$$

Add -1 and 4 to get 3.

$$-18t^{2}+6t+3$$

Expand

$$3+6t-18t^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3t-1\right)^{2}.

$$-\left(9t^{2}-6t+1\right)-\left(3t+2\right)\left(3t-2\right)$$

To find the opposite of 9t^{2}-6t+1, find the opposite of each term.

$$-9t^{2}+6t-1-\left(3t+2\right)\left(3t-2\right)$$

Consider \left(3t+2\right)\left(3t-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$-9t^{2}+6t-1-\left(\left(3t\right)^{2}-4\right)$$

Expand \left(3t\right)^{2}.

$$-9t^{2}+6t-1-\left(3^{2}t^{2}-4\right)$$

Calculate 3 to the power of 2 and get 9.

$$-9t^{2}+6t-1-\left(9t^{2}-4\right)$$

To find the opposite of 9t^{2}-4, find the opposite of each term.

$$-9t^{2}+6t-1-9t^{2}+4$$

Combine -9t^{2} and -9t^{2} to get -18t^{2}.

$$-18t^{2}+6t-1+4$$

Add -1 and 4 to get 3.

$$-18t^{2}+6t+3$$