Solve f(x)=21[(x-1/6)^2-49/36]

$$f ( x ) = 21 [ ( x - \frac { 1 } { 6 } ) ^ { 2 } - \frac { 49 } { 36 } ]$$

Evaluate

$$7\left(3x-4\right)\left(x+1\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{6}\right)^{2}.

$$21\left(x^{2}-\frac{1}{3}x+\frac{1}{36}-\frac{49}{36}\right)$$

Subtract \frac{49}{36} from \frac{1}{36} to get -\frac{4}{3}.

$$21\left(x^{2}-\frac{1}{3}x-\frac{4}{3}\right)$$

Use the distributive property to multiply 21 by x^{2}-\frac{1}{3}x-\frac{4}{3}.

$$21x^{2}-7x-28$$

Expand

$$21x^{2}-7x-28$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{1}{6}\right)^{2}.

$$21\left(x^{2}-\frac{1}{3}x+\frac{1}{36}-\frac{49}{36}\right)$$

Subtract \frac{49}{36} from \frac{1}{36} to get -\frac{4}{3}.

$$21\left(x^{2}-\frac{1}{3}x-\frac{4}{3}\right)$$

Use the distributive property to multiply 21 by x^{2}-\frac{1}{3}x-\frac{4}{3}.

$$21x^{2}-7x-28$$