STAAR Review Equivalent Questions

The zeros are -\frac{3}{2} and \frac{27}{2} because f(x)=(2x+3)(2x+27)

The zeros are \pm\frac{9}{2} because f(x)=(2x-9)(2x+9)

The zeros are \frac{3}{4} and -27 because f(x)=(4x-3)(x-27)

The only zero is \frac{9}{2} because f(x)=(2x-9)^{2}

x^{\frac{17}{7}}

x^{\frac{6}{14}}

x^{\frac{5}{7}}

x^{\frac{6}{7}}

4w - 12

4w + 12

w + 3

w - 3

25\sqrt{24}

6\sqrt{10}

24\sqrt{25}

10\sqrt{6}

6a^{2}+11a-10

17a - 10

6a^{2}-10

6a^{2}+19a+10

12g(2h-g+3)

12g(2h-12g+18)

6g(4h-12g+18)

6g(4h-2g+3)

B=\frac{2S}{Ph}

B=\frac{S-2}{Ph}

B=\frac{SPh}{2}

B=\frac{S-Ph}{2}

2n^{2}-n-28

3n+3

n-28

2n^{2}-15n-28

c^{6}d^{18}

c^{6}d^{9}

c^{4}d^{9}

c^{4}d^{18}

x = -3

x = -5

x = 3

x = 5

4\times10^{-3}

4\times10^{-6}

4\times10^{12}

4\times10^6

11rt+rw-10tw

11rt-7rw-10tw

11rt+rw-6tw

11rt-7rw-6tw

The only zero, 6, can be found when 0 - (x - 6)(x - 6).

The only zeros, -18 and 18, can be found when 0 = (x + 18)(x - 18)

The only zero, 18, can be found when 0 = (x - 18)(x - 18).

The only zeros, -6 and 6, can be found when 0 = (x + 6)(x - 6)

(9m - 20)(4m + 5)

2(2m - 5)(9m+10)

4(3m-5)^2

4(3m - 5)(3m + 5)

k(x)=(x+1)(x-15)

k(x)=(x+15)(x-12)

k(x)=(x+3)(x-5)

k(x)=(x+5)(x-3)