The first 3 are examples of the difference of 2 squares so you use the identity
a^2 - b^2 = (a + b)(a - b)
x^2 - 49 = 0
so (x + 7)(x - 7) = 0
so either x + 7 = 0 or x - 7 = 0
giving x = -7 and 7.
Number 7 reduces to 3x^2 =12, x^2 = 4 so x = +/- 2
Number 8 take out GCf (d) to give
d(d - 2) = 0 so d = 0 , 2
9 and 10 are more difficult to factor
you use the 'ac' method Google it to get more details
2x^2 - 5x + 2
multiply first coefficient by the constant at the end
that is 2 * 2 = 4
Now we want 2 numbers which when multiplied give + 4 and when added give - 5:- -1 and -4 seem promising so we write the equation as:-
2x^2 - 4x - x + 2 = 0
now factor by grouping
2x(x - 2) - 1(x - 2) = 0
(x - 2) is common so
(2x - 1)(x - 2) = 0
and 2x - 1 = 0 or x - 2 = 0 and now you can find x.
The last example is solved in the same way.
Answer:
The third expression: (2x)^3
Step-by-step explanation:
Plug in 3 for x in each expression
2x^3
2(3)^3
Use PEMDAS
2 (3 × 3 × 3)
2 (9 × 3)
2 (27)
54
2x^3 + 5
(The last expression told us 2x^3 = 54 so feel free to use this information to make life easier)
54 + 5
59
(2x)^3
(2 × 3)^3
Use PEMDAS
6^3
6 × 6 × 6
36 × 6
216
(x - 1)^3
(3 - 1)^3
Use PEMDAS
2^3
2 × 2 × 2
4 × 2
8
216 is the greatest value so the third expression is the answer
The first part of the second line, she left the -5 there. The correct work and solution should be this:
5(2x-1)-3x=5x+9
<span><span>7x</span>−5</span>=<span><span>5x</span>+<span>9
</span></span>2x-5=9
2x=14
x=7
