Quadratic and Polynomial Relationships
Help Questions
GRE Quantitative Reasoning › Quadratic and Polynomial Relationships
If $h(x)=x^2+2x-15$, which of the following is a factor of $h(x)$?
$x+15$
$x-15$
$x+5$
$x+3$
$x-3$
Explanation
This question tests finding factors of a quadratic polynomial. To find factors of h(x) = x² + 2x - 15, we need to factor the expression. We look for two numbers that multiply to -15 and add to 2; these are 5 and -3. Therefore, h(x) = (x + 5)(x - 3), giving us factors (x + 5) and (x - 3). Choice C correctly identifies (x + 5) as a factor. A common mistake is to confuse the signs in factoring, potentially selecting (x - 5) or mixing up which values work.
If $x$ is a real number and $x^2-5x+6=0$, which of the following is a possible value of $x$?
$\frac{3}{2}$
$-2$
$6$
$2$
$-3$
Explanation
This question tests quadratic polynomial relationships by asking us to solve a quadratic equation. To find possible values of x, we need to factor the quadratic expression x² - 5x + 6 = 0. We look for two numbers that multiply to 6 and add to -5; these are -2 and -3. Therefore, x² - 5x + 6 = (x - 2)(x - 3) = 0, giving us x = 2 or x = 3. The correct answer is B, which gives x = 2 as a possible value. A common error would be to confuse the signs when factoring, leading to incorrect values like x = -2.
Which of the following is equivalent to the polynomial expression $x^2-9$?
$(x-9)(x+1)$
$x(x-9)$
$(x-3)(x+3)$
$(x-3)^2$
$(x+3)^2$
Explanation
This question tests polynomial factorization, specifically recognizing a difference of squares pattern. The expression x² - 9 follows the pattern a² - b² = (a + b)(a - b), where a = x and b = 3. Applying this formula, we get x² - 9 = x² - 3² = (x + 3)(x - 3). Therefore, the correct factorization is (x - 3)(x + 3), which matches choice C. A common mistake is to factor this as (x - 3)², which would expand to x² - 6x + 9, not x² - 9.
Which of the following statements about the quadratic function $g(x)=-2x^2+8x+3$ is true?
The parabola opens downward because the coefficient of $x^2$ is negative.
The parabola opens upward because the coefficient of $x^2$ is negative.
The vertex occurs at $x=8$.
The parabola has a $y$-intercept of $8$.
The parabola has no real $x$-intercepts.
Explanation
This question tests understanding of quadratic function properties, particularly the effect of the leading coefficient. In the function g(x) = -2x² + 8x + 3, the coefficient of x² is -2, which is negative. When the leading coefficient of a quadratic is negative, the parabola opens downward, making choice B correct. The y-intercept occurs when x = 0, giving g(0) = 3, not 8 as choice C suggests. Choice A incorrectly states that the parabola opens upward despite acknowledging the negative coefficient.
For the polynomial $f(x)=x^2-4x-5$, what is the value of $f(5)-f(-1)$?
$-20$
$-10$
$0$
$10$
$20$
Explanation
This question tests evaluating polynomial functions and computing differences. To find f(5) - f(-1), we must evaluate f(x) = x² - 4x - 5 at both x = 5 and x = -1. For f(5): f(5) = 5² - 4(5) - 5 = 25 - 20 - 5 = 0. For f(-1): f(-1) = (-1)² - 4(-1) - 5 = 1 + 4 - 5 = 0. Therefore, f(5) - f(-1) = 0 - 0 = 0, giving us answer A. A common error would be to make arithmetic mistakes when substituting negative values, potentially leading to incorrect answers like -20 or 20.
A polynomial is defined by $p(x)=(x-1)(x+4)(x-2)$. Which of the following statements is true?
$p(x)$ has a double zero at $x=1$.
$p(x)$ has zeros at $x=-1,4,2$.
$p(x)$ has a zero at $x=0$.
$p(x)$ has exactly two real zeros.
$p(x)$ has zeros at $x=1,-4,2$.
Explanation
This question tests understanding of polynomial zeros from factored form. When a polynomial is written as p(x) = (x - 1)(x + 4)(x - 2), the zeros occur when any factor equals zero. Setting each factor to zero: x - 1 = 0 gives x = 1; x + 4 = 0 gives x = -4; x - 2 = 0 gives x = 2. Therefore, the polynomial has zeros at x = 1, -4, and 2, which matches choice D. Choice A incorrectly identifies the zero from (x + 4) as x = 4 instead of x = -4, a common sign error when working with factored forms.
The height (in meters) of a ball $t$ seconds after it is thrown is given by $H(t)=-5t^2+20t+1$. At what time $t$ does the ball reach its maximum height?
$1$
$2$
$10$
$5$
$4$
Explanation
This question tests finding the maximum of a quadratic function in a real-world context. For a quadratic function in the form H(t) = at² + bt + c with a < 0, the maximum occurs at t = -b/(2a). In H(t) = -5t² + 20t + 1, we have a = -5 and b = 20. Therefore, the maximum occurs at t = -20/(2(-5)) = -20/(-10) = 2 seconds. The answer is B. A common error would be to use the wrong formula or make sign errors in the calculation, potentially getting t = 4 or another incorrect value.
If $q(x)=x^3-3x^2-4x+12$, which of the following is the value of $q(3)$?
$0$
$-6$
$6$
$-12$
$12$
Explanation
This question tests evaluating a cubic polynomial at a specific value. To find q(3) where q(x) = x³ - 3x² - 4x + 12, we substitute x = 3 into the expression. Computing step by step: q(3) = 3³ - 3(3²) - 4(3) + 12 = 27 - 3(9) - 12 + 12 = 27 - 27 - 12 + 12 = 0. Therefore, q(3) = 0, making A the correct answer. A common error would be arithmetic mistakes in computing powers or handling negative terms, potentially leading to answers like 6 or -6.
Which of the following statements is true about the quadratic $y=ax^2+bx+c$ if $a>0$ and $c<0$?
The vertex must lie on the $x$-axis.
The $y$-intercept is below the $x$-axis.
The graph must cross the $x$-axis at exactly one point.
The graph has no real $x$-intercepts.
The graph opens downward.
Explanation
This question explores properties of quadratic graphs based on coefficients. For y = ax² + bx + c with a>0 and c<0, the parabola opens upward, and the y-intercept is c, which is negative. This means the graph crosses the y-axis below the x-axis. The sign of a determines the direction, and c directly gives the y-intercept. This justifies choice D as true. A distractor like choice A fails because a>0 means it opens upward, not downward, a sign misinterpretation. Another, like choice B, is not necessarily true since the discriminant could allow real intercepts.
Which of the following statements is true about the polynomial $q(x)=-2x^2+7x-3$?
The graph opens downward because the leading coefficient is negative.
The $y$-intercept is $7$.
The axis of symmetry is $x=7$.
The graph must have no real zeros because the leading coefficient is negative.
The graph opens upward because the constant term is negative.
Explanation
This question examines properties of quadratic polynomials from coefficients. For q(x) = -2x² + 7x - 3, the leading coefficient -2 is negative, so the parabola opens downward. This determines the end behavior for large |x|. The y-intercept is -3, not 7, and axis is x = -7/(2*(-2)) = 7/4. This justifies choice B as true. A distractor like choice D fails because the y-intercept is the constant term -3, possibly confusing with the linear coefficient. Another, like choice A, errs on the sign of the leading coefficient.