Answers to precalculus math questions

1.) Factor the trinomial.
4(a + b)^2 + 7(a + b) − 2

2.) Factor the polynomial.
x^2 + 8x + 16 - 25y^2

3.) Find all values of b for which the trinomial can be factored.
x^2 + bx + 135

4.) Completely factor the expression.
(x^2 + 63)^2 – 256x^2

5.) Multiply or find the special product.
[(x – 6) + y]^2

1.
You can use substitution if you find that it makes it easier.
Let u = a + b.
4u^2 + 7u - 2 =
(4u - 1)(u + 2)
Now substitute a+b back for u.
(4(a + b) - 1)((a + b) + 2)
(4a + 4b - 1)(a + b + 2)

2.
(x+4)^2 - 25y^2
This is the difference of squares. a^2 - b^2 = (a+b)(a-b). In this case, a = x+4 and b = 5y.
(x + 4 + 5y)(x + 4 - 5y)

3.
x^2 + bx + 135
If 135 is factored into 1 and 135, then b = 135 + 1 = 136.
If 135 is factored into 3 and 45, then b = 3 + 45 = 48.
If 135 is factored into 5 and 27, then b = 5 + 27 = 32.
If 135 is factored into 9 and 15, then b = 9 + 15 = 24.
So the possible values for b are 24, 32, 48, and 136.

4.
(x^2 + 63)^2 - 256x^2
Difference of squares again. a = x^2 + 63, and b = 16x.
(x^2 + 63 + 16x)(x^2 + 63 - 16x)
Rearrange the terms and conitnue factoring.
(x^2 + 16x + 63)(x^2 - 16x + 63)
(x + 9)(x + 7)(x - 9)(x - 7)

5.
[(x - 6) + y]^2 =
(x - 6)^2 + 2(x-6)y + y^2 =
x^2 - 12x + 36 + 2xy - 12y + y^2

Let a + b = x

4x^2 + 7x - 2 = (4x - 1)(x + 2)

Put back a + b

4(a + b - 1)(a + b + 2)


x^2 + 9x + 16 - 25y^2

Two quadratic terms indicate a circle, ellipse or hyperbola. Expect to complete the square.


x^2 + b + 135 has real roots if the discriminant, b^2 - 4ac >= 0. a = 1, b = b, c = 135


In problem 4, you have the difference of two squares: a^2 - b^2 = (a + b)(a - b)