This question asks for the factored form of (y = $x^2$ + 8x + 7). Factor by finding numbers that multiply to 7 and add to 8, which are 1 and 7, so (y = (x + 1)(x + 7)). Verify by expanding: $(x^2$ + 7x + x + 7 = $x^2$ + 8x + 7). The quadratic formula gives roots -1 and -7, confirming factors. Avoid errors like choosing numbers that multiply incorrectly, such as -1 and -7 which add to -8. When factoring quadratics with positive constant and leading coefficient 1, look for positive factors if b>0.

This question asks us to find the solutions to the quadratic equation $2x^2-5x-3=0$. We can solve this by factoring: we need two numbers that multiply to $(2)(-3)=-6$ and add to $-5$. These numbers are $-6$ and $1$, so we rewrite as $2x^2-6x+x-3=0$, then factor by grouping: $2x(x-3)+1(x-3)=0$, giving us $(2x+1)(x-3)=0$. Setting each factor to zero: $2x+1=0$ gives $x=-\frac{1}{2}$, and $x-3=0$ gives $x=3$. A common error is making arithmetic mistakes when factoring or solving for x. When you see a quadratic equation on the PSAT, first check if it factors nicely before using the quadratic formula.

This question asks for the time when the ball reaches maximum height, given the height function $h(t)=-16t^2+48t+5$. For a quadratic function in the form $at^2+bt+c$, the maximum (when $a<0$) occurs at $t=-\frac{b}{2a}$. Here, $a=-16$ and $b=48$, so the maximum occurs at $t=-\frac{48}{2(-16)}=-\frac{48}{-32}=\frac{48}{32}=\frac{3}{2}$. We can verify this is a maximum since the coefficient of $t^2$ is negative, meaning the parabola opens downward. A common error is forgetting the negative sign in the formula or making arithmetic mistakes with fractions. For projectile motion problems, remember that the vertex formula gives you the time of maximum height directly.

This question seeks the time (t) when a ball reaches maximum height, modeled by (h(t) = $-5t^2$ + 20t + 1). For quadratics, the vertex gives the maximum (since a<0), at (t = -rac{b}{2a} = -rac{20}{2(-5)} = rac{20}{10} = 2). Completing the square: (h(t) = $-5(t^2$ - 4t) + 1 = -5(t - $2)^2$ + 21), confirming peak at t=2. The quadratic formula isn't needed for vertex time. A common mistake is using positive a or forgetting the negative sign. In projectile problems, remember the vertex formula directly gives the time to max height.

This question asks to solve $(2x^2$ + 3x - 2 = 0) and select the solutions. Factor: find numbers for (2x + a)(x + b) where a*b=-2, and cross terms give 3x; try (2x - 1)(x + 2) = $2x^2$ + 4x - x - 2 = $2x^2$ + 3x - 2, yes, so x=1/2 or x=-2. Quadratic formula: (x = rac{-3 pm $\sqrt{9 + 16}$}{4} = rac{-3 pm 5}{4}), giving (rac{2}{4}=0.5) and (rac{-8}{4}=-2). Factoring works well with small integers; formula confirms. Common error: incorrect factoring signs. Verify by plugging solutions back in.

The question asks when the ball reaches maximum height given $h(t)=-5t^2+20t+1$. For a quadratic in the form $at^2+bt+c$, the maximum (when $a<0$) occurs at $t = -\frac{b}{2a}$. Here, $a=-5$ and $b=20$, so $t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2$ seconds. We can verify this is a maximum since the coefficient of $t^2$ is negative (-5), making the parabola open downward. A common mistake is forgetting the negative sign in the formula or confusing when to use maximum vs minimum. For time-based problems, always check that your answer makes physical sense (positive time).

The question asks for the axis of symmetry of y = -2x² + 8x - 3. The axis is the vertical line x = -b/(2a), with a = -2, b = 8, so x = -8/(2*(-2)) = -8/-4 = 2. This is the line through the vertex, confirmed by vertex form after completing the square: y = -2(x² - 4x) - 3 = -2(x² - 4x + 4 - 4) - 3 = -2(x - 2)² + 8 - 3 = -2(x - 2)² + 5. Mistaking it for y-values like the y-intercept -3 or another number is common. Confusing with positive b sign can lead to x = -2. Always use the formula x = -b/(2a) for quick identification in graphing questions.

This question requires finding the equation of the axis of symmetry for a parabola that opens upward and crosses the x-axis at (x = -1) and (x = 3). The axis of symmetry for a parabola is the vertical line passing through the midpoint of the roots, calculated as (x = rac{-1 + 3}{2} = 1). You can also think of it as (x = -rac{b}{2a}) in the standard form, but here the roots directly give the symmetry line. Using the quadratic formula isn't necessary since the roots are provided, but it would confirm the same axis. A key error is confusing the axis with the y-intercept or miscalculating the midpoint, such as averaging incorrectly to get 2. For parabolas given by roots, always use the midpoint formula as a quick strategy to find the axis without full equation derivation.

The vertex form $y=a(x-h)^2+k$ directly shows the vertex at $(h,k)$. In the equation $y=2(x-3)^2-8$, we identify $h=3$ and $k=-8$, so the vertex is at $(3,-8)$. Note that the sign inside the parentheses is opposite to the x-coordinate: $(x-3)$ means the vertex has x-coordinate positive 3. A common error is thinking $(x-3)$ gives vertex at $x=-3$, but remember the form is $(x-h)$, not $(x+h)$. When given vertex form, you can immediately read off the vertex without any calculations.

This question asks for the value of x that maximizes the area of a garden, given A(x) = x(24 - 2x) = 24x - $2x^2$. This is a quadratic that opens downward (a = -2), so the maximum occurs at the vertex, x = -rac{b}{2a} = -rac{24}{2(-2)} = rac{24}{4} = 6. Completing the square: A(x) = $-2(x^2$ - 12x) = -2(x - $6)^2$ + 72, confirming x=6. A key error is treating it as minimizing instead of maximizing or misexpanding the area formula. Another mistake is solving for roots rather than the vertex. In optimization problems, convert to quadratic form and use the vertex formula to find the extremum efficiently.