PRACTICE QUESTIONS ON SOLVING OPERATIONS WITH POLYNOMIALS FOR SAT

Problem 1 :

Which of the following gives the solution set for the polynomial equation below?

𝑥² − 11𝑥 + 19 = −5

A) {-3, -8}     B) {3, 8}     C) {-3, 8}      D) {3, -8}

Solution :

𝑥² − 11𝑥 + 19 = −5

𝑥² − 11𝑥 + 19 + 5 = 0

𝑥² − 11𝑥 + 24 = 0

(x - 8)(x - 3) = 0

x = 8 and x = 3

The solution is x = {3, 8}.

So, option B is correct.

Problem 2 :

Which of the following is a solution for the equation below?

7𝑟² − 14𝑟 = −7

A) -1     B) 0     C) 1       D) 7

Solution :

7𝑟² − 14𝑟 = −7

Adding 7 on both sides

7𝑟² − 14𝑟 + 7 = 0

𝑟² − 2𝑟 + 1 = 0

(r - 1)(r - 1) = 0

r = 1

So, the solution is 1.

Problem 3 :

Which of the following is equivalent to the expression above?

Solution :

Problem 4 :

If x > 3, which expression is equivalent to

Solution :

Problem 5 :

If

(ax + 2)(bx + 7) = 15x² + cx + 14

for all values of x, and a+b = 8, what are the two possible values for c ?

A) 3 and 5        B) 6 and 35       C) 10 and 21      D) 31 and 41

Solution :

(ax + 2)(bx + 7) = 15x² + cx + 14

abx² + 7ax + 2bx + 14 = 15x² + cx + 14

abx² + (7a + 2b)x + 14 = 15x² + cx + 14

ab = 15, 7a + 2b = c

ab = 15 -----(1)

7a + 2b = c -----(2)

a + b = 8 ----(3)

From (3), b = 8 - a

Applying the value of b in (1), we get

a(8 - a) = 15

8a - a² = 15

a² -8a + 15 = 0

(a - 3)(a - 5) = 0

a = 3 and a = 5

Applying the value of a in b = 8 - a

Applying the value of a and b in (2), we get

So, the solutions are 31 and 41.

Problem 6 :

If t > 0 and t² - 4 = 0, what is the value of t ?

Solution :

t² - 4 = 0

t² = 4

t = √4

t = ±2

t = 2 and t = -2

Since t > 0, the solution is 2.

Problem 7 :

h(t) = −4.9t² + 25t

The equation above expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 meters per second. After approximately how many seconds will the ball hit the ground?

A) 3.5      B) 4.0      C) 4.5      D) 5.0

Solution :

h(t) = −4.9t² + 25t

Since we find the time taken when it reaches the ground.

−4.9t² + 25t = 0

-t((4.9t - 25) = 0

t = 0 and 4.9t - 25 = 0

t = 25/4.9

t = 5.10.

So, the answer is option D.

Problem 8 :

For a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x ) ?

A) x − 5 is a factor of p (x)         B) x − 2 is a factor of p(x)

C) x + 2 is a factor of p(x)

D) The remainder when p(x) is divided by x − 3 is −2.

Solution :

Let p(x) be a polynomial, If (x - a) is a factor, then p(a) = 0

Here p(3) = -2 ≠ 0

Then 3, that x - 3 is not a factor. It must be the divisor. When we divide the polynomial p(x) by x - 3, we get the remainder -2. So, option D is correct.

Problem 9 :

For what value of x is the function h above undefined?

Solution :

At x = 3, the function is not defined.

Problem 10 :

x² - 34x + c = 0

In the given equation, c is a constant. The equation has no real solutions if c > n . What is the least possible value of n?

Solution :

Since the given quadratic equation is not having real solution

b² - 4ac < 0

a = 1, b = -34 and c = c

(-34)² - 4(1) (c) < 0

1156 - 4c < 0

Subtracting 1156 on both sides.

-4c < -1156

Dividing by negative on both sides.

c > 1156/4

c > 289

Given that c > n

n = 289

So, the least value of n is 289.