Polynomial Equations and Quadratic Expressions – Advanced Math (No Calculator) for the SAT (50 Questions)

Solution

We look for two numbers that multiply to 6 (the constant term) and add up to 5 (the coefficient of x). Those numbers are 2 and 3. Hence:

x² + 5x + 6 = (x + 2)(x + 3)

This is one of the most basic forms of factoring a quadratic: a monic quadratic expression with small integer coefficients.

Answer: (x + 2)(x + 3).

Solution

For a quadratic of the form ax² + bx + c, we look for two numbers that multiply to a*c = 2*10 = 20 and add to b = 9. The two numbers that fit are 5 and 4. We then rewrite the middle term 9x as 5x + 4x:

2x² + 9x + 10 = 2x² + 5x + 4x + 10

Factor by grouping:
(2x² + 5x) + (4x + 10)
= x(2x + 5) + 2(2x + 5)
= (2x + 5)(x + 2)

Thus, the complete factorization is (2x + 5)(x + 2).

Solution

First, factor the left-hand side if possible:
x² - x - 6 = (x - 3)(x + 2)

By the Zero Product Property, (x - 3)(x + 2) = 0 implies x - 3 = 0 or x + 2 = 0. Thus, x = 3 or x = -2.

Answer: x = 3 or x = -2.

Solution

x² - 16 is a difference of squares. Recall that a² - b² = (a - b)(a + b). Here, a = x and b = 4:
x² - 16 = (x - 4)(x + 4)

Answer: (x - 4)(x + 4).

Solution

This is also a difference of squares, with a = 2x and b = 5:
4x² - 25 = (2x)² - 5² = (2x - 5)(2x + 5)

Answer: (2x - 5)(2x + 5).

Solution

First, factor out the greatest common factor (GCF). Both terms share 3x:
3x² - 12x = 3x(x - 4)

Once the GCF is factored out, the resulting binomial (x - 4) has no further factorization (over the integers).

Answer: 3x(x - 4).

Solution

Separate into two groups and factor each:
(x³ + 2x²) + (3x + 6)
= x²(x + 2) + 3(x + 2)
Now factor out (x + 2):
= (x + 2)(x² + 3)

Note that x² + 3 does not factor further over the real numbers (it has no real roots).

Answer: (x + 2)(x² + 3).

Solution

We look for two numbers that multiply to 6*(-3) = -18 and add to -7. The pair is -9 and 2:
-9 + 2 = -7
-9 * 2 = -18

Rewrite the middle term -7x as -9x + 2x:
6x² - 7x - 3 = 6x² - 9x + 2x - 3
Group:
(6x² - 9x) + (2x - 3)
= 3x(2x - 3) + 1(2x - 3)
= (2x - 3)(3x + 1)

Answer: (2x - 3)(3x + 1).

Solution

This might be factorable, but if not immediately obvious, use the quadratic formula. Let’s see if it factors. We want two numbers that multiply to 2*(-1) = -2 and add to 1. The pair is 2 and -1. So we rewrite:
2x² + x - 1 = 2x² + 2x - x - 1
= 2x(x + 1) - 1(x + 1)
= (2x - 1)(x + 1)

Set each factor to zero:
2x - 1 = 0 → x = 1/2
x + 1 = 0 → x = -1

Answer: x = 1/2 or x = -1.

Solution

x³ + 8 is a sum of cubes. Recall the sum of cubes factorization: a³ + b³ = (a + b)(a² - ab + b²). Here, a = x and b = 2, because 8 = 2³.
x³ + 8 = (x + 2)(x² - 2x + 4)

Answer: (x + 2)(x² - 2x + 4).

Solution

x³ - 27 is a difference of cubes. The formula for a³ - b³ is (a - b)(a² + ab + b²). Here, a = x and b = 3:
x³ - 27 = (x - 3)(x² + 3x + 9)

Answer: (x - 3)(x² + 3x + 9).

Solution

Factor or simply notice x² = 4. Factoring approach:
x² - 4 = (x - 2)(x + 2) = 0
Solutions: x = 2 or x = -2

Answer: x = 2 or x = -2.

Solution

Look for the greatest common factor first:
4x³ - 8x² + 4x = 4x(x² - 2x + 1)

Notice x² - 2x + 1 is a perfect square trinomial: (x - 1)². So we have:
4x(x - 1)²

Answer: 4x(x - 1)².

Solution

The quadratic formula for ax² + bx + c = 0 is x = [-b ± √(b² - 4ac)] / (2a). Here, a = 3, b = -1, c = -4. Plugging in:
x = [ -(-1) ± √((-1)² - 4*3*(-4)) ] / (2*3)
= (1 ± √(1 + 48)) / 6
= (1 ± √49) / 6
= (1 ± 7) / 6

Thus, x = (1 + 7)/6 = 8/6 = 4/3, or x = (1 - 7)/6 = -6/6 = -1.

Answer: x = 4/3 or x = -1.

Solution

Expand and rewrite as a quadratic equation:
x² + x = 12
x² + x - 12 = 0

Factor if possible. We need two numbers that multiply to -12 and add to 1: 4 and -3. So:
x² + x - 12 = (x + 4)(x - 3)
= 0

Thus, x = -4 or x = 3.

Answer: x = -4 or x = 3.

Solution

x² + 4x + 4 is a perfect square trinomial: (x + 2)². So set (x + 2)² = 0. This implies x + 2 = 0, or x = -2. The root has multiplicity 2.

Answer: x = -2 (double root).

Solution

Move 1 to the other side: x² = -1. Over the real numbers, there is no real solution because a square of a real number cannot be negative. The solutions, if we allow complex numbers, are x = ±i. However, for the SAT (and real solutions), there is no real solution.

Answer: No real solutions.

Solution

Group terms: (2x³ - 2x²) + (4x - 4). Factor each group:
2x²(x - 1) + 4(x - 1)
Now factor out (x - 1):
(x - 1)(2x² + 4)
We can further factor out a 2 from (2x² + 4):
(x - 1)(2)(x² + 2)
So final:
2(x - 1)(x² + 2)

Answer: 2(x - 1)(x² + 2).

Solution

Divide both sides by 9:
x² = 16/9
So x = ±(4/3).

Answer: x = 4/3 or x = -4/3.

Solution

Factor out the greatest common factor first. All terms share 3x:
6x³ - 9x² - 24x = 3x(2x² - 3x - 8)

Now factor the quadratic 2x² - 3x - 8. We need two numbers that multiply to 2*(-8) = -16 and add to -3. Those numbers are -4 and 1. Rewrite:
2x² - 3x - 8 = 2x² - 4x + x - 8
= 2x(x - 2) + 1(x - 8) … but that doesn’t look right; let's do it carefully. Actually, let's systematically do grouping:

We want -4 + 1 = -3, so we try:
2x² + x - 4x - 8
= x(2x + 1) - 4(x + 2).
That also doesn't match precisely. Let’s refine. Since we want to factor 2x² - 3x - 8, let's test possible factor forms: (2x + a)(x + b). The product must be 2x² - 3x - 8. The possible integer pairs for a*b = -8 are:
(1, -8), (-1, 8), (2, -4), (-2, 4).
We look for a sum that yields -3 in the x coefficient. Checking systematically:
(2x + 1)(x - 8) gives 2x² -16x + x -8 = 2x² -15x -8, not correct.
(2x - 1)(x + 8) gives 2x² +16x - x -8 = 2x² +15x -8, also not correct.
(2x + 2)(x - 4) = 2x² -8x +2x -8 = 2x² -6x -8, that has -6 for the x coefficient, not -3.
(2x - 2)(x + 4) = 2x² +8x -2x -8 = 2x² +6x -8, that’s +6, not -3.
(2x + 4)(x - 2) = 2x² -4x +4x -8 = 2x² 0x -8, not correct.
(2x - 4)(x + 2) = 2x² +4x -4x -8 = 2x² -8, not correct.
(2x + 8)(x - 1) = 2x² -2x +8x -8 = 2x² +6x -8, also not correct.
(2x - 8)(x + 1) = 2x² +2x -8x -8 = 2x² -6x -8, not correct.

Possibly, it might factor to (2x + 1)(x - 4) if we check that: 2x² - 8x + x - 4 = 2x² -7x -4. Not correct.

Let's see if it factors nicely at all or if we made a sign error. We want 2x² - 3x - 8. The discriminant is b² - 4ac = (-3)² - 4(2)(-8) = 9 + 64 = 73. Since 73 is prime and not a perfect square, 2x² - 3x - 8 does not factor over the integers.
Therefore, the factorization might only be partial with a GCF. So the final factorization is:
3x(2x² - 3x - 8)

Since 2x² - 3x - 8 doesn’t factor over the integers, we stop there.

Answer: 3x(2x² - 3x - 8).

Solution

Let’s see if it factors nicely. We want two numbers that multiply to 2*2 = 4 and sum to 5. Those numbers are 4 and 1.
2x² + 5x + 2 = 2x² + 4x + x + 2
= 2x(x + 2) + 1(x + 2)
= (2x + 1)(x + 2)

Set each factor to zero:
2x + 1 = 0 → x = -1/2
x + 2 = 0 → x = -2

Answer: x = -1/2 or x = -2.

Solution

Distribute the minus sign over the second set of parentheses:
3x² + 2x - 1 - x² + 4x - 5
Combine like terms:
(3x² - x²) + (2x + 4x) + (-1 - 5)
= 2x² + 6x - 6

Answer: 2x² + 6x - 6.

Solution

Expand the left-hand side:
(x - 1)(x + 3) = x² + 3x - x - 3 = x² + 2x - 3
So the equation is x² + 2x - 3 = 2
Subtract 2 from both sides:
x² + 2x - 5 = 0

Now solve x² + 2x - 5 = 0 using the quadratic formula, a = 1, b = 2, c = -5:
x = [-2 ± √(2² - 4*1*(-5))] / (2*1)
= [-2 ± √(4 + 20)] / 2
= [-2 ± √24] / 2
= [-2 ± 2√6] / 2
= -1 ± √6

Answer: x = -1 + √6 or x = -1 - √6.

Solution

We want two numbers a and b that multiply to 3 and add to 4. Those numbers are 1 and 3. So x² + 4x + 3 = (x + 1)(x + 3).

Answer: (a, b) = (1, 3).

Solution

Notice there is a common factor of 9:
9x² - 9 = 9(x² - 1)
Then factor the difference of squares inside:
x² - 1 = (x - 1)(x + 1)
So overall:
9(x - 1)(x + 1)

Answer: 9(x - 1)(x + 1).

Solution

Let’s see if this is a perfect square trinomial. 4x² = (2x)², 9 = 3², and the middle term 12x might match 2 * 2x * 3 = 12x. Indeed, that fits. So:
4x² + 12x + 9 = (2x + 3)²

Answer: (2x + 3)².

Solution

Group in pairs:
(2x³ + 10x²) + (3x + 15)
Factor out common terms from each group:
2x²(x + 5) + 3(x + 5)
Now factor out (x + 5):
(x + 5)(2x² + 3)

Since 2x² + 3 does not factor further over the integers, we are done.

Answer: (x + 5)(2x² + 3).

Solution

Factor the left side. We want two numbers that multiply to 6 and add to -7. The pair is -6 and -1 (since -6 * -1 = 6 and -6 + -1 = -7). So:
x² - 7x + 6 = (x - 6)(x - 1) = 0
The solutions are x = 6 or x = 1.

Answer: x = 6 or x = 1.

Solution

a = 4, b = -4, c = -3. Plug into x = [-b ± √(b² - 4ac)] / (2a):
x = [ -(-4) ± √((-4)² - 4*4*(-3)) ] / (2*4)
= (4 ± √(16 + 48)) / 8
= (4 ± √64) / 8
= (4 ± 8) / 8

Thus, x = (4 + 8)/8 = 12/8 = 3/2, or x = (4 - 8)/8 = -4/8 = -1/2.

Answer: x = 3/2 or x = -1/2.

Solution

We look for two numbers that multiply to 6*(-2) = -12 and add to 1 (the coefficient of x). That pair is 4 and -3. Rewrite the middle term:
6x² + x - 2 = 6x² + 4x - 3x - 2
Group:
(6x² + 4x) + (-3x - 2)
= 2x(3x + 2) -1(3x + 2)
= (3x + 2)(2x - 1)

Answer: (3x + 2)(2x - 1).

Solution

Take square roots on both sides:
x + 2 = ±7
Case 1: x + 2 = 7 → x = 5
Case 2: x + 2 = -7 → x = -9

Answer: x = 5 or x = -9.

Solution

Combine like terms:
(3x² + x²) + (-4x + 5x) + (1 - 2)
= 4x² + x - 1

Answer: 4x² + x - 1.

Solution

First, take out the greatest common factor. All terms share 4:
12x² + 4x - 16 = 4(3x² + x - 4)
Now factor 3x² + x - 4 if possible. We want two numbers that multiply to 3*(-4) = -12 and add to 1. Those numbers are 4 and -3. Rewrite:
3x² + x - 4 = 3x² + 4x - 3x - 4
= x(3x + 4) - 1(3x + 4)
= (3x + 4)(x - 1)

Hence, the full factorization is:
4(3x + 4)(x - 1)

Answer: 4(3x + 4)(x - 1).

Solution

If x = 2 is a root, then f(2) = 0:
3(2)² - (2) + k = 0
3(4) - 2 + k = 0
12 - 2 + k = 0
10 + k = 0
k = -10

Answer: k = -10.

Solution

8x³ + 64 can first factor out 8:
8(x³ + 8)
Then x³ + 8 is a sum of cubes. x³ + 2³ = (x + 2)(x² - 2x + 4).
So the final factorization is:
8(x + 2)(x² - 2x + 4)

Answer: 8(x + 2)(x² - 2x + 4).

Solution

Recognize that x² - 6x + 9 is a perfect square trinomial: (x - 3)². So set (x - 3)² = 0, which implies x - 3 = 0, so x = 3.
This root has multiplicity 2.

Answer: x = 3 (double root).

Solution

Expand the left-hand side:
(x - 2)(3x + 1) = 3x² + x - 6x - 2 = 3x² - 5x - 2
So the equation becomes:
3x² - 5x - 2 = 5
Subtract 5 from both sides:
3x² - 5x - 7 = 0

Now we attempt to solve 3x² - 5x - 7 = 0 using the quadratic formula with a=3, b=-5, c=-7:
x = [ -(-5) ± √((-5)² - 4*3*(-7)) ] / (2*3)
= (5 ± √(25 + 84)) / 6
= (5 ± √109) / 6

Since 109 is not a perfect square, these are the exact solutions.

Answer: x = (5 + √109)/6 or x = (5 - √109)/6.

Solution

Use the quadratic formula with a=1, b=2, c=-1:
x = [ -2 ± √(2² - 4*1*(-1)) ] / (2*1)
= ( -2 ± √(4 + 4) ) / 2
= ( -2 ± √8 ) / 2
= ( -2 ± 2√2 ) / 2
= -1 ± √2

Answer: x = -1 + √2 or x = -1 - √2.

Solution

Group the terms:
(6x³ + 9x²) + (2x + 3)
Factor out common factors in each group:
3x²(2x + 3) + 1(2x + 3)
Now factor out (2x + 3):
(2x + 3)(3x² + 1)

Answer: (2x + 3)(3x² + 1).

Solution

4x² = 1
x² = 1/4
x = ±1/2

Alternatively, factor 4x² - 1 as (2x - 1)(2x + 1) = 0. Solutions are x = 1/2 or x = -1/2.

Answer: x = 1/2 or x = -1/2.

Solution

First, factor out the greatest common factor. All terms share 4x:
4x³ + 16x² - 20x = 4x(x² + 4x - 5)
Now factor x² + 4x - 5 if it factors nicely. We want two numbers that multiply to -5 and add to 4. That pair is 5 and -1. So:
x² + 4x - 5 = (x + 5)(x - 1)

Therefore:
4x(x + 5)(x - 1)

Answer: 4x(x + 5)(x - 1).

Solution

Rearrange to get a standard quadratic:
5x² - 6x - 1 = 0
Use the quadratic formula: a=5, b=-6, c=-1.
x = [ -(-6) ± √((-6)² - 4*5*(-1)) ] / (2*5)
= (6 ± √(36 + 20)) / 10
= (6 ± √56) / 10
= (6 ± √(4*14)) / 10
= (6 ± 2√14) / 10
= (3 ± √14) / 5

Answer: x = (3 + √14)/5 or x = (3 - √14)/5.

Solution

Factor f(x) = x² - 3x + 2:
x² - 3x + 2 = (x - 1)(x - 2)
Setting it equal to 0 gives x = 1 or x = 2.

Answer: x = 1 or x = 2.

Solution

First, take out the greatest common factor of 3:
3x² + 12x + 12 = 3(x² + 4x + 4)
Inside, x² + 4x + 4 is a perfect square: (x + 2)².
So the expression becomes:
3(x + 2)²

Answer: 3(x + 2)².

Solution

By the Zero Product Property, if x(x - 4)(x + 1) = 0, then one of the factors must be zero:
x = 0, or (x - 4) = 0 → x = 4, or (x + 1) = 0 → x = -1.

Answer: x = 0, x = 4, or x = -1.

Solution

Expand: 2x(x + 3) = 2x² + 6x. So the equation is 2x² + 6x = 5.
Move 5 to the left side:
2x² + 6x - 5 = 0

We can try factoring or use the quadratic formula. Let’s attempt factoring. We want two numbers that multiply to 2*(-5) = -10 and add to 6. Those numbers are 10 and -4? That sums to 6 indeed. So rewrite:
2x² + 10x - 4x - 5
= 2x(x + 5) - 1(4x + 5)? That’s not correct. Let’s carefully group:
2x² + 10x - 4x - 5
= 2x(x + 5) - (4x + 5)
This doesn’t share a factor. So maybe factoring is not straightforward or leads to a fractional factor. Let’s do the quadratic formula:
a=2, b=6, c=-5.
x = [ -6 ± √(6² - 4*2*(-5)) ] / (2*2)
= ( -6 ± √(36 + 40) ) / 4
= ( -6 ± √76 ) / 4
= ( -6 ± 2√19 ) / 4
= ( -3 ± √19 ) / 2

Answer: x = (-3 + √19)/2 or x = (-3 - √19)/2.

Solution

x⁴ - 16 can be seen as a difference of squares: (x²)² - 4². So:
x⁴ - 16 = (x² - 4)(x² + 4)
Next, factor x² - 4 further: that’s (x - 2)(x + 2). Meanwhile, x² + 4 does not factor over the real numbers (it can factor over the complex domain as (x - 2i)(x + 2i)).

So over the reals, the complete factorization is:
(x - 2)(x + 2)(x² + 4)

Answer: (x - 2)(x + 2)(x² + 4).

Solution

Expand: 3x(2x - 5) = 6x² - 15x. So 6x² - 15x = 4.
Move 4 to the left:
6x² - 15x - 4 = 0

Use the quadratic formula: a=6, b=-15, c=-4.
x = [ -(-15) ± √((-15)² - 4*6*(-4)) ] / (2*6)
= (15 ± √(225 + 96)) / 12
= (15 ± √321) / 12
√321 = √(9* 35.666...)? Not a perfect square. So we keep it as √321.

Answer: x = [15 + √321]/12 or x = [15 - √321]/12.

Solution

Group terms:
(5x³ - 10x²) + (-x + 2)
= 5x²(x - 2) -1(x - 2)
Factor out (x - 2):
(x - 2)(5x² - 1)

Answer: (x - 2)(5x² - 1).

Solution

Observe that 4x² + 4x + 1 = (2x + 1)². Let’s confirm by expansion: (2x + 1)(2x + 1) = 4x² + 2x + 2x + 1 = 4x² + 4x + 1.
Setting (2x + 1)² = 0 → 2x + 1 = 0 → x = -1/2.

Answer: x = -1/2 (double root).