Study Quadratics And Polynomials in ACT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Factor 4 x 2 − 9 4x^2 - 9 4 x 2 − 9 . What is the expression? Answer: ( 2 x − 3 ) ( 2 x + 3 ) (2x - 3)(2x + 3) ( 2 x − 3 ) ( 2 x + 3 ) . Difference of squares: ( 2 x ) 2 − 3 2 (2x)^2 - 3^2 ( 2 x ) 2 − 3 2 .
Flashcard 2: Factor x 2 + 5 x + 6 x^2 + 5x + 6 x 2 + 5 x + 6 completely. Answer: ( x + 2 ) ( x + 3 ) (x + 2)(x + 3) ( x + 2 ) ( x + 3 ) . Find factors of 6 that add to 5: 2 and 3.
Flashcard 3: Identify the leading coefficient in 4 x 3 − x 2 + 5 4x^3 - x^2 + 5 4 x 3 − x 2 + 5 . Answer:
The coefficient of the term with the highest degree.
Flashcard 4: For what discriminant values does a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 have real and equal roots? Answer: Discriminant b 2 − 4 a c = 0 b^2 - 4ac = 0 b 2 − 4 a c = 0 . Zero discriminant means one repeated real solution.
Flashcard 5: Find f ( 2 ) f(2) f ( 2 ) for f ( x ) = x 2 − 3 x + 2 f(x) = x^2 - 3x + 2 f ( x ) = x 2 − 3 x + 2 . Answer: f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 . Substitute x = 2 x=2 x = 2 : ( 2 ) 2 − 3 ( 2 ) + 2 = 4 − 6 + 2 = 0 (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0 ( 2 ) 2 − 3 ( 2 ) + 2 = 4 − 6 + 2 = 0 .
Flashcard 6: Factor x 2 − 1 x^2 - 1 x 2 − 1 . Answer: ( x + 1 ) ( x − 1 ) (x+1)(x-1) ( x + 1 ) ( x − 1 ) . Difference of squares: x 2 − 1 2 = ( x + 1 ) ( x − 1 ) x^2 - 1^2 = (x+1)(x-1) x 2 − 1 2 = ( x + 1 ) ( x − 1 ) .
Flashcard 7: Solve for x x x : 2 x 2 + 7 x + 3 = 0 2x^2+7x+3=0 2 x 2 + 7 x + 3 = 0 . Answer: x = − 3 x=-3 x = − 3 or x = − 1 2 x=-\frac{1}{2} x = − 2 1 . Factor as ( 2 x + 1 ) ( x + 3 ) = 0 (2x+1)(x+3)=0 ( 2 x + 1 ) ( x + 3 ) = 0 and solve each factor.
Flashcard 8: State the difference of squares factorization formula. Answer: a 2 − b 2 = ( a − b ) ( a + b ) a^2-b^2=(a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b ) . Special factoring pattern for expressions with squared terms.
Flashcard 9: State the factor theorem relating f ( a ) f(a) f ( a ) to the factor ( x − a ) (x-a) ( x − a ) . Answer: ( x − a ) (x-a) ( x − a ) is a factor iff f ( a ) = 0 f(a)=0 f ( a ) = 0 . A polynomial has ( x − a ) (x-a) ( x − a ) as a factor if and only if f ( a ) = 0 f(a)=0 f ( a ) = 0 .
Flashcard 10: What are the roots of x 2 − 9 x^2 - 9 x 2 − 9 ? Answer: x = 3 , − 3 x = 3, -3 x = 3 , − 3 . Difference of squares: x 2 − 9 = ( x − 3 ) ( x + 3 ) = 0 x^2 - 9 = (x-3)(x+3) = 0 x 2 − 9 = ( x − 3 ) ( x + 3 ) = 0 .
Flashcard 11: What does Δ = 0 \Delta=0 Δ = 0 indicate about the real solutions of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 ? Answer: One real double solution. Zero discriminant means the parabola touches the x-axis at exactly one point.
Flashcard 12: What is the vertex form of a quadratic function? Answer: y = a ( x − h ) 2 + k y = a(x - h)^2 + k y = a ( x − h ) 2 + k . Form that shows vertex ( h , k ) (h,k) ( h , k ) directly.
Flashcard 13: State the binomial theorem. Answer: ( x + y ) n = ∑ k = 0 n ( n k ) x n − k y k (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k ( x + y ) n = ∑ k = 0 n ( k n ) x n − k y k . Formula for expanding ( x + y ) n (x+y)^n ( x + y ) n using binomial coefficients.
Flashcard 14: Factor x 2 − 9 x^2 - 9 x 2 − 9 completely. Answer: ( x − 3 ) ( x + 3 ) (x - 3)(x + 3) ( x − 3 ) ( x + 3 ) . Difference of squares: a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b ) .
Flashcard 15: What is the result of multiplying x ( x + 1 ) x(x+1) x ( x + 1 ) ? Answer: x 2 + x x^2 + x x 2 + x . Distribute: x ( x ) + x ( 1 ) = x 2 + x x(x) + x(1) = x^2 + x x ( x ) + x ( 1 ) = x 2 + x .
Flashcard 16: Find the discriminant of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . Answer: b 2 − 4 a c b^2 - 4ac b 2 − 4 a c . Expression under the square root in the quadratic formula.
Flashcard 17: Identify the axis of symmetry for y = a x 2 + b x + c y = ax^2 + bx + c y = a x 2 + b x + c . Answer: x = − b 2 a x = \frac{-b}{2a} x = 2 a − b . X-coordinate of the parabola's vertex and line of symmetry.
Flashcard 18: State the difference of squares factorization formula. Answer: a 2 − b 2 = ( a − b ) ( a + b ) a^2-b^2=(a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b ) . Special factoring pattern for expressions with squared terms.
Flashcard 19: What is the y y y -intercept of y = a x 2 + b x + c y=ax^2+bx+c y = a x 2 + b x + c ? Answer: ( 0 , c ) (0,c) ( 0 , c ) . Set x = 0 x=0 x = 0 in the equation to find where the parabola crosses the y-axis.
Flashcard 20: State the definition of the degree of a polynomial. Answer: Largest exponent of x x x with nonzero coefficient. The highest power of the variable with a non-zero coefficient.
Flashcard 21: Simplify 2 x 2 − 4 x + 2 2x^2 - 4x + 2 2 x 2 − 4 x + 2 . What is the expression? Answer: 2 ( x 2 − 2 x + 1 ) 2(x^2 - 2x + 1) 2 ( x 2 − 2 x + 1 ) . Factor out common factor of 2.
Flashcard 22: State the quadratic formula for solutions to a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 . Answer: x = − b ± b 2 − 4 a c 2 a x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . Derived by completing the square on the general quadratic equation.
Flashcard 23: Which term is the quadratic term in 5 x 2 − 3 x + 8 5x^2 - 3x + 8 5 x 2 − 3 x + 8 ? Answer: Quadratic term is 5 x 2 5x^2 5 x 2 . Term with x 2 x^2 x 2 (degree 2).
Flashcard 24: What is the standard form of a quadratic equation? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . General form with a ≠ 0 a \neq 0 a = 0 for quadratic equations.
Flashcard 25: State Vieta's formulas for a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 with roots r 1 , r 2 r_1,r_2 r 1 , r 2 . Answer: r 1 + r 2 = − b a r_1+r_2=-\frac{b}{a} r 1 + r 2 = − a b and r 1 r 2 = c a r_1r_2=\frac{c}{a} r 1 r 2 = a c . Relationships between coefficients and roots without solving the equation.
Flashcard 26: Evaluate x 2 − 4 x + 4 x^2 - 4x + 4 x 2 − 4 x + 4 for x = 2 x = 2 x = 2 . Answer: Result is 0 0 0 . Substitute x = 2 x=2 x = 2 : ( 2 ) 2 − 4 ( 2 ) + 4 = 4 − 8 + 4 = 0 (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0 ( 2 ) 2 − 4 ( 2 ) + 4 = 4 − 8 + 4 = 0 .
Flashcard 27: Solve for x x x : x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 Answer: x = 2 , 3 x = 2, 3 x = 2 , 3 . Factor as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 , so x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 .
Flashcard 28: What is the degree of the polynomial 7 x 5 − 4 x 3 + 2 7x^5 - 4x^3 + 2 7 x 5 − 4 x 3 + 2 ? Answer:
The highest power of the variable in the polynomial.
Flashcard 29: For what discriminant values does a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 have complex roots? Answer: Discriminant b 2 − 4 a c < 0 b^2 - 4ac < 0 b 2 − 4 a c < 0 . Negative discriminant means no real solutions.
Flashcard 30: Identify the constant term in 3 x 2 − 5 x + 7 3x^2 - 5x + 7 3 x 2 − 5 x + 7 . Answer:
The term without a variable (degree 0).
Flashcard 31: Simplify ( x + 1 ) ( x − 1 ) (x + 1)(x - 1) ( x + 1 ) ( x − 1 ) . What is the expression? Answer: x 2 − 1 x^2 - 1 x 2 − 1 . Difference of squares formula.
Flashcard 32: What is the y-intercept of y = 3 x 2 + 2 x − 4 y = 3x^2 + 2x - 4 y = 3 x 2 + 2 x − 4 ? Answer: Y-intercept is − 4 -4 − 4 . Y-intercept occurs when x = 0 x=0 x = 0 , giving y = c y=c y = c .
Flashcard 33: Factor x 2 − 4 x + 4 x^2 - 4x + 4 x 2 − 4 x + 4 . Answer: ( x − 2 ) 2 (x-2)^2 ( x − 2 ) 2 . Perfect square trinomial: ( x − 2 ) 2 = x 2 − 4 x + 4 (x-2)^2 = x^2 - 4x + 4 ( x − 2 ) 2 = x 2 − 4 x + 4 .
Flashcard 34: What is the vertex of y = ( x − 2 ) 2 + 3 y = (x-2)^2 + 3 y = ( x − 2 ) 2 + 3 ? Answer: Vertex is ( 2 , 3 ) (2, 3) ( 2 , 3 ) . In vertex form, ( h , k ) (h,k) ( h , k ) gives the vertex coordinates.
Flashcard 35: State the zero-product property used when solving ( A ) ( B ) = 0 (A)(B)=0 ( A ) ( B ) = 0 . Answer: A = 0 A=0 A = 0 or B = 0 B=0 B = 0 . If a product equals zero, at least one factor must equal zero.
Flashcard 36: Find f ( 2 ) f(2) f ( 2 ) for f ( x ) = x 2 − 3 x + 2 f(x) = x^2 - 3x + 2 f ( x ) = x 2 − 3 x + 2 . Answer: f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 . Substitute x = 2 x=2 x = 2 : ( 2 ) 2 − 3 ( 2 ) + 2 = 4 − 6 + 2 = 0 (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0 ( 2 ) 2 − 3 ( 2 ) + 2 = 4 − 6 + 2 = 0 .
Flashcard 37: Find the discriminant of 3 x 2 − 4 x + 5 = 0 3x^2-4x+5=0 3 x 2 − 4 x + 5 = 0 . Answer: Δ = − 44 \Delta=-44 Δ = − 44 . Calculate Δ = ( − 4 ) 2 − 4 ( 3 ) ( 5 ) = 16 − 60 = − 44 \Delta=(-4)^2-4(3)(5)=16-60=-44 Δ = ( − 4 ) 2 − 4 ( 3 ) ( 5 ) = 16 − 60 = − 44 .
Flashcard 38: For what discriminant values does a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 have real and distinct roots? Answer: Discriminant b 2 − 4 a c > 0 b^2 - 4ac > 0 b 2 − 4 a c > 0 . Positive discriminant means two different real solutions.
Flashcard 39: Factor x 2 − 4 x + 4 x^2 - 4x + 4 x 2 − 4 x + 4 . Answer: ( x − 2 ) 2 (x-2)^2 ( x − 2 ) 2 . Perfect square trinomial: ( x − 2 ) 2 = x 2 − 4 x + 4 (x-2)^2 = x^2 - 4x + 4 ( x − 2 ) 2 = x 2 − 4 x + 4 .
Flashcard 40: Find the vertex of y = x 2 − 4 x + 4 y = x^2 - 4x + 4 y = x 2 − 4 x + 4 . Answer: ( 2 , 0 ) (2, 0) ( 2 , 0 ) . Complete the square or use x = − b 2 a = 2 x = -\frac{b}{2a} = 2 x = − 2 a b = 2 , y = 0 y = 0 y = 0 .
Flashcard 41: State the standard form of a quadratic function in x x x . Answer: f ( x ) = a x 2 + b x + c f(x)=ax^2+bx+c f ( x ) = a x 2 + b x + c , with a ≠ 0 a\ne 0 a = 0 . General form where a a a determines parabola direction and cannot be zero.
Flashcard 42: Find the remainder when x 3 − 2 x 2 + x − 1 x^3 - 2x^2 + x - 1 x 3 − 2 x 2 + x − 1 is divided by x − 1 x - 1 x − 1 . Answer: Remainder is − 1 -1 − 1 . By remainder theorem: f ( 1 ) = 1 − 2 + 1 − 1 = − 1 f(1) = 1 - 2 + 1 - 1 = -1 f ( 1 ) = 1 − 2 + 1 − 1 = − 1 .
Flashcard 43: What is the discriminant of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 ? Answer: Δ = b 2 − 4 a c \Delta=b^2-4ac Δ = b 2 − 4 a c . Expression under the square root in the quadratic formula.
Flashcard 44: Find the greatest common factor of 2 x 3 2x^3 2 x 3 and 4 x 2 4x^2 4 x 2 . Answer: 2 x 2 2x^2 2 x 2 . Factor out the highest power common to both terms.
Flashcard 45: What are the roots of x 2 − 9 x^2 - 9 x 2 − 9 ? Answer: x = 3 , − 3 x = 3, -3 x = 3 , − 3 . Difference of squares: x 2 − 9 = ( x − 3 ) ( x + 3 ) = 0 x^2 - 9 = (x-3)(x+3) = 0 x 2 − 9 = ( x − 3 ) ( x + 3 ) = 0 .
Flashcard 46: If f ( x ) = x 2 − 1 f(x)=x^2-1 f ( x ) = x 2 − 1 , what are the zeros of f ( x ) f(x) f ( x ) ? Answer: x = − 1 x=-1 x = − 1 and x = 1 x=1 x = 1 . Set f ( x ) = 0 f(x)=0 f ( x ) = 0 : x 2 − 1 = 0 x^2-1=0 x 2 − 1 = 0 , so ( x − 1 ) ( x + 1 ) = 0 (x-1)(x+1)=0 ( x − 1 ) ( x + 1 ) = 0 .
Flashcard 47: State the factor theorem. Answer: If f ( c ) = 0 f(c) = 0 f ( c ) = 0 , then ( x − c ) (x - c) ( x − c ) is a factor of f ( x ) f(x) f ( x ) . If c c c is a root, then ( x − c ) (x-c) ( x − c ) divides the polynomial.
Flashcard 48: Identify the vertex x x x -coordinate of y = 2 x 2 − 8 x + 1 y=2x^2-8x+1 y = 2 x 2 − 8 x + 1 . Answer: x = 2 x=2 x = 2 . Use x = − b 2 a = − − 8 2 ( 2 ) = 2 x=-\frac{b}{2a}=-\frac{-8}{2(2)}=2 x = − 2 a b = − 2 ( 2 ) − 8 = 2 .
Flashcard 49: Subtract ( 3 x 2 + 2 x ) (3x^2 + 2x) ( 3 x 2 + 2 x ) from ( 5 x 2 + x + 4 ) (5x^2 + x + 4) ( 5 x 2 + x + 4 ) . Answer: 2 x 2 − x + 4 2x^2 - x + 4 2 x 2 − x + 4 . Subtract each term: ( 5 − 3 ) x 2 + ( 1 − 2 ) x + ( 4 − 0 ) (5-3)x^2 + (1-2)x + (4-0) ( 5 − 3 ) x 2 + ( 1 − 2 ) x + ( 4 − 0 ) .
Flashcard 50: What does Δ = 0 \Delta=0 Δ = 0 indicate about the real solutions of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 ? Answer: One real double solution. Zero discriminant means the parabola touches the x-axis at exactly one point.
Flashcard 51: Evaluate x 2 − 4 x + 4 x^2 - 4x + 4 x 2 − 4 x + 4 for x = 2 x = 2 x = 2 . Answer: Result is 0 0 0 . Substitute x = 2 x=2 x = 2 : ( 2 ) 2 − 4 ( 2 ) + 4 = 4 − 8 + 4 = 0 (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0 ( 2 ) 2 − 4 ( 2 ) + 4 = 4 − 8 + 4 = 0 .
Flashcard 52: Factor x 2 − 9 x^2 - 9 x 2 − 9 completely. Answer: ( x − 3 ) ( x + 3 ) (x - 3)(x + 3) ( x − 3 ) ( x + 3 ) . Difference of squares: a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b ) .
Flashcard 53: Simplify 2 x 2 − 4 x + 2 2x^2 - 4x + 2 2 x 2 − 4 x + 2 . What is the expression? Answer: 2 ( x 2 − 2 x + 1 ) 2(x^2 - 2x + 1) 2 ( x 2 − 2 x + 1 ) . Factor out common factor of 2.
Flashcard 54: What is the linear term in the polynomial 7 x 2 + 2 x − 1 7x^2 + 2x - 1 7 x 2 + 2 x − 1 ? Answer: Linear term is 2 x 2x 2 x . Term with x x x (degree 1).
Flashcard 55: State the remainder theorem for dividing f ( x ) f(x) f ( x ) by ( x − a ) (x-a) ( x − a ) . Answer: Remainder = f ( a ) =f(a) = f ( a ) . The remainder equals the function value at the divisor's root.
Flashcard 56: State the axis of symmetry for y = a x 2 + b x + c y=ax^2+bx+c y = a x 2 + b x + c . Answer: x = − b 2 a x=-\frac{b}{2a} x = − 2 a b . The x-coordinate where the parabola reaches its maximum or minimum.
Flashcard 57: What does the discriminant indicate about the roots? Answer: Number and nature of roots. Positive: two real roots; zero: one root; negative: no real roots.
Flashcard 58: Identify the vertex form of a quadratic function. Answer: y = a ( x − h ) 2 + k y = a(x-h)^2 + k y = a ( x − h ) 2 + k . Form showing vertex ( h , k ) (h,k) ( h , k ) and vertical stretch/compression a a a .
Flashcard 59: In y = a ( x − h ) 2 + k y=a(x-h)^2+k y = a ( x − h ) 2 + k , what are the vertex coordinates? Answer: Vertex = ( h , k ) =(h,k) = ( h , k ) . The vertex is at the point ( h , k ) (h,k) ( h , k ) in this form.
Flashcard 60: Find the remainder when x 3 − 2 x 2 + x − 1 x^3 - 2x^2 + x - 1 x 3 − 2 x 2 + x − 1 is divided by x − 1 x - 1 x − 1 . Answer: Remainder is − 1 -1 − 1 . By remainder theorem: f ( 1 ) = 1 − 2 + 1 − 1 = − 1 f(1) = 1 - 2 + 1 - 1 = -1 f ( 1 ) = 1 − 2 + 1 − 1 = − 1 .
Flashcard 61: Simplify x 3 − x 2 + x − 1 x^3 - x^2 + x - 1 x 3 − x 2 + x − 1 for x = 1 x = 1 x = 1 . Answer: Result is 0 0 0 . Substitute x = 1 x=1 x = 1 : 1 − 1 + 1 − 1 = 0 1 - 1 + 1 - 1 = 0 1 − 1 + 1 − 1 = 0 .
Flashcard 62: Determine the roots of x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 . Answer: Roots are x = 2 x = 2 x = 2 and x = 3 x = 3 x = 3 . Factor as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 , so x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 .
Flashcard 63: What is the standard form of a quadratic equation? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . General form where a ≠ 0 a ≠ 0 a = 0 and equation equals zero.
Flashcard 64: How many real solutions does x 2 + 4 x + 5 = 0 x^2+4x+5=0 x 2 + 4 x + 5 = 0 have? Answer: Zero real solutions. Discriminant Δ = 16 − 20 = − 4 < 0 \Delta=16-20=-4<0 Δ = 16 − 20 = − 4 < 0 , so no real solutions.
Flashcard 65: Simplify: ( x + 2 ) ( x − 5 ) (x+2)(x-5) ( x + 2 ) ( x − 5 ) . Answer: x 2 − 3 x − 10 x^2-3x-10 x 2 − 3 x − 10 . Use FOIL: ( x + 2 ) ( x − 5 ) = x 2 − 5 x + 2 x − 10 (x+2)(x-5)=x^2-5x+2x-10 ( x + 2 ) ( x − 5 ) = x 2 − 5 x + 2 x − 10 .
Flashcard 66: What is the product of ( x + 2 ) (x+2) ( x + 2 ) and ( x − 3 ) (x-3) ( x − 3 ) ? Answer: x 2 − x − 6 x^2 - x - 6 x 2 − x − 6 . Use FOIL: x 2 − 3 x + 2 x − 6 = x 2 − x − 6 x^2 - 3x + 2x - 6 = x^2 - x - 6 x 2 − 3 x + 2 x − 6 = x 2 − x − 6 .
Flashcard 67: State the factored form of a quadratic with roots r 1 r_1 r 1 and r 2 r_2 r 2 . Answer: a ( x − r 1 ) ( x − r 2 ) a(x-r_1)(x-r_2) a ( x − r 1 ) ( x − r 2 ) . Form showing the quadratic as a product using its roots or zeros.
Flashcard 68: State Vieta's formulas for a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 with roots r 1 , r 2 r_1,r_2 r 1 , r 2 . Answer: r 1 + r 2 = − b a r_1+r_2=-\frac{b}{a} r 1 + r 2 = − a b and r 1 r 2 = c a r_1r_2=\frac{c}{a} r 1 r 2 = a c . Relationships between coefficients and roots without solving the equation.
Flashcard 69: What is the constant term in the polynomial x 3 + 4 x 2 − 2 x + 6 x^3 + 4x^2 - 2x + 6 x 3 + 4 x 2 − 2 x + 6 ? Answer: Constant term is 6. Term with no variable (degree 0).
Flashcard 70: What is the discriminant of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 ? Answer: Δ = b 2 − 4 a c \Delta=b^2-4ac Δ = b 2 − 4 a c . Expression under the square root in the quadratic formula.
Flashcard 71: Solve for x x x : x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 Answer: x = 2 , 3 x = 2, 3 x = 2 , 3 . Factor as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3) = 0 ( x − 2 ) ( x − 3 ) = 0 , so x = 2 x = 2 x = 2 or x = 3 x = 3 x = 3 .
Flashcard 72: Determine if x = − 2 x = -2 x = − 2 is a root of x 3 + 2 x 2 − 5 x^3 + 2x^2 - 5 x 3 + 2 x 2 − 5 . Answer: Not a root. Substitute x = − 2 x=-2 x = − 2 : ( − 2 ) 3 + 2 ( − 2 ) 2 − 5 = − 8 + 8 − 5 = − 5 ≠ 0 (-2)^3 + 2(-2)^2 - 5 = -8 + 8 - 5 = -5 \neq 0 ( − 2 ) 3 + 2 ( − 2 ) 2 − 5 = − 8 + 8 − 5 = − 5 = 0 .
Flashcard 73: What does Δ < 0 \Delta<0 Δ < 0 indicate about the solutions of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 ? Answer: No real solutions (two complex solutions). Negative discriminant means the parabola doesn't cross the x-axis.
Flashcard 74: What is the y-intercept of y = 3 x 2 + 2 x − 4 y = 3x^2 + 2x - 4 y = 3 x 2 + 2 x − 4 ? Answer: Y-intercept is − 4 -4 − 4 . Y-intercept occurs when x = 0 x=0 x = 0 , giving y = c y=c y = c .
Flashcard 75: What is the result of multiplying x ( x + 1 ) x(x+1) x ( x + 1 ) ? Answer: x 2 + x x^2 + x x 2 + x . Distribute: x ( x ) + x ( 1 ) = x 2 + x x(x) + x(1) = x^2 + x x ( x ) + x ( 1 ) = x 2 + x .
Flashcard 76: What is the sum of the coefficients in 2 x 2 − 3 x + 5 2x^2 - 3x + 5 2 x 2 − 3 x + 5 ? Answer:
Add all coefficients: 2 + ( − 3 ) + 5 = 4 2 + (-3) + 5 = 4 2 + ( − 3 ) + 5 = 4 .
Flashcard 77: Factor completely: x 2 + 10 x + 25 x^2+10x+25 x 2 + 10 x + 25 . Answer: ( x + 5 ) 2 (x+5)^2 ( x + 5 ) 2 . Perfect square trinomial: a 2 + 2 a b + b 2 = ( a + b ) 2 a^2+2ab+b^2=(a+b)^2 a 2 + 2 ab + b 2 = ( a + b ) 2 .
Flashcard 78: State the zero-product property used when solving ( A ) ( B ) = 0 (A)(B)=0 ( A ) ( B ) = 0 . Answer: A = 0 A=0 A = 0 or B = 0 B=0 B = 0 . If a product equals zero, at least one factor must equal zero.
Flashcard 79: Evaluate f ( 2 ) f(2) f ( 2 ) for f ( x ) = x 3 − 4 x + 1 f(x)=x^3-4x+1 f ( x ) = x 3 − 4 x + 1 . Answer: 1 1 1 . Substitute: f ( 2 ) = 2 3 − 4 ( 2 ) + 1 = 8 − 8 + 1 = 1 f(2)=2^3-4(2)+1=8-8+1=1 f ( 2 ) = 2 3 − 4 ( 2 ) + 1 = 8 − 8 + 1 = 1 .
Flashcard 80: Find the vertex of y = x 2 − 6 x + 5 y=x^2-6x+5 y = x 2 − 6 x + 5 . Answer: ( 3 , − 4 ) (3,-4) ( 3 , − 4 ) . Complete the square or use vertex formula x = − b 2 a = 3 x=-\frac{b}{2a}=3 x = − 2 a b = 3 , then y = 5 − 9 = − 4 y=5-9=-4 y = 5 − 9 = − 4 .
Flashcard 81: State the maximum number of real zeros a degree n n n polynomial can have. Answer: At most n n n real zeros. A polynomial of degree n n n can cross the x-axis at most n n n times.
Flashcard 82: What is the degree of the polynomial 7 x 5 − 4 x 3 + 2 7x^5 - 4x^3 + 2 7 x 5 − 4 x 3 + 2 ? Answer:
The highest power of the variable in the polynomial.
Flashcard 83: Identify the cubic term in 4 x 3 + 3 x 2 + x + 5 4x^3 + 3x^2 + x + 5 4 x 3 + 3 x 2 + x + 5 . Answer: Cubic term is 4 x 3 4x^3 4 x 3 . Term with x 3 x^3 x 3 (degree 3).
Flashcard 84: Expand and simplify: ( x − 4 ) 2 (x-4)^2 ( x − 4 ) 2 . Answer: x 2 − 8 x + 16 x^2-8x+16 x 2 − 8 x + 16 . Use the formula ( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2 = a^2 - 2ab + b^2 ( a − b ) 2 = a 2 − 2 ab + b 2 .
Flashcard 85: Find the vertex of y = x 2 − 6 x + 5 y=x^2-6x+5 y = x 2 − 6 x + 5 . Answer: ( 3 , − 4 ) (3,-4) ( 3 , − 4 ) . Complete the square or use vertex formula x = − b 2 a = 3 x=-\frac{b}{2a}=3 x = − 2 a b = 3 , then y = 5 − 9 = − 4 y=5-9=-4 y = 5 − 9 = − 4 .
Flashcard 86: Identify the degree of x 3 + 2 x 2 − 5 x + 1 x^3 + 2x^2 - 5x + 1 x 3 + 2 x 2 − 5 x + 1 . Answer:
The highest power of x x x in the polynomial is 3.
Flashcard 87: State the factored form of a quadratic with roots r 1 r_1 r 1 and r 2 r_2 r 2 . Answer: a ( x − r 1 ) ( x − r 2 ) a(x-r_1)(x-r_2) a ( x − r 1 ) ( x − r 2 ) . Form showing the quadratic as a product using its roots or zeros.
Flashcard 88: What is the coefficient of x x x in 5 x 2 + 3 x − 7 5x^2 + 3x - 7 5 x 2 + 3 x − 7 ? Answer:
The numerical factor multiplying the x x x term.
Flashcard 89: Evaluate f ( 2 ) f(2) f ( 2 ) for f ( x ) = x 3 − 4 x + 1 f(x)=x^3-4x+1 f ( x ) = x 3 − 4 x + 1 . Answer: 1 1 1 . Substitute: f ( 2 ) = 2 3 − 4 ( 2 ) + 1 = 8 − 8 + 1 = 1 f(2)=2^3-4(2)+1=8-8+1=1 f ( 2 ) = 2 3 − 4 ( 2 ) + 1 = 8 − 8 + 1 = 1 .
Flashcard 90: What is the vertex of y = ( x − 2 ) 2 + 3 y = (x-2)^2 + 3 y = ( x − 2 ) 2 + 3 ? Answer: Vertex is ( 2 , 3 ) (2, 3) ( 2 , 3 ) . In vertex form, ( h , k ) (h,k) ( h , k ) gives the vertex coordinates.
Flashcard 91: What is the sum of the solutions to 2 x 2 − 8 x = 0 2x^2 - 8x = 0 2 x 2 − 8 x = 0 ? Answer: Sum is 4 4 4 . Factor as 2 x ( x − 4 ) = 0 2x(x-4) = 0 2 x ( x − 4 ) = 0 , roots are x = 0 , 4 x=0,4 x = 0 , 4 ; sum is 0 + 4 = 4 0+4=4 0 + 4 = 4 .
Flashcard 92: Determine if x = − 2 x = -2 x = − 2 is a root of x 3 + 2 x 2 − 5 x^3 + 2x^2 - 5 x 3 + 2 x 2 − 5 . Answer: Not a root. Substitute x = − 2 x=-2 x = − 2 : ( − 2 ) 3 + 2 ( − 2 ) 2 − 5 = − 8 + 8 − 5 = − 5 ≠ 0 (-2)^3 + 2(-2)^2 - 5 = -8 + 8 - 5 = -5 \neq 0 ( − 2 ) 3 + 2 ( − 2 ) 2 − 5 = − 8 + 8 − 5 = − 5 = 0 .
Flashcard 93: For what discriminant values does a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 have real and equal roots? Answer: Discriminant b 2 − 4 a c = 0 b^2 - 4ac = 0 b 2 − 4 a c = 0 . Zero discriminant means one repeated real solution.
Flashcard 94: Simplify ( 2 x − 3 ) 2 (2x - 3)^2 ( 2 x − 3 ) 2 . What is the expression? Answer: 4 x 2 − 12 x + 9 4x^2 - 12x + 9 4 x 2 − 12 x + 9 . Perfect square trinomial: ( a − b ) 2 = a 2 − 2 a b + b 2 (a-b)^2 = a^2 - 2ab + b^2 ( a − b ) 2 = a 2 − 2 ab + b 2 .
Flashcard 95: Determine the x-intercepts of y = x 2 − 4 y = x^2 - 4 y = x 2 − 4 . Answer: x = 2 x = 2 x = 2 and x = − 2 x = -2 x = − 2 . Set y = 0 y=0 y = 0 and solve: x 2 − 4 = 0 x^2 - 4 = 0 x 2 − 4 = 0 .
Flashcard 96: State the quadratic formula. Answer: x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . Formula to solve any quadratic equation a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 .
Flashcard 97: What is the product of the roots of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 ? Answer: c a \frac{c}{a} a c . By Vieta's formulas for quadratic equations.
Flashcard 98: Factor completely: x 2 − 9 x^2-9 x 2 − 9 . Answer: ( x − 3 ) ( x + 3 ) (x-3)(x+3) ( x − 3 ) ( x + 3 ) . Difference of squares pattern: a 2 − b 2 = ( a − b ) ( a + b ) a^2-b^2=(a-b)(a+b) a 2 − b 2 = ( a − b ) ( a + b ) .
Flashcard 99: Find the zero of the polynomial x − 4 x - 4 x − 4 . Answer: x = 4 x = 4 x = 4 . Set the polynomial equal to zero: x − 4 = 0 x - 4 = 0 x − 4 = 0 .
Flashcard 100: What is the sum of the roots of a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 ? Answer: − b a -\frac{b}{a} − a b . By Vieta's formulas for quadratic equations.