Study Polynomial Equations in PSAT 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: What does the discriminant value b 2 − 4 a c > 0 b^2-4ac>0 b 2 − 4 a c > 0 imply about real solutions? Answer: Two distinct real solutions. Positive discriminant yields two different real roots.
Flashcard 2: What is the sum of the roots of x 2 − 7 x + 10 = 0 x^2-7x+10=0 x 2 − 7 x + 10 = 0 ? Answer: 7 7 7 . For a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , sum of roots equals − b a -\frac{b}{a} − a b .
Flashcard 3: Evaluate P ( 2 ) P(2) P ( 2 ) for P ( x ) = x 3 − 4 x P(x)=x^3-4x P ( x ) = x 3 − 4 x : what is P ( 2 ) P(2) P ( 2 ) ? Answer: 0 0 0 . P ( 2 ) = 2 3 − 4 ( 2 ) = 8 − 8 = 0 P(2)=2^3-4(2)=8-8=0 P ( 2 ) = 2 3 − 4 ( 2 ) = 8 − 8 = 0 .
Flashcard 4: What is the factored form of 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: 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 5: Factor completely: x 2 + 6 x + 9 x^2+6x+9 x 2 + 6 x + 9 . Answer: ( x + 3 ) 2 (x+3)^2 ( x + 3 ) 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 6: Identify the solutions of x 2 − 9 = 0 x^2-9=0 x 2 − 9 = 0 . Answer: x = ± 3 x=\pm 3 x = ± 3 . Factor as ( x − 3 ) ( x + 3 ) = 0 (x-3)(x+3)=0 ( x − 3 ) ( x + 3 ) = 0 , giving x = 3 x=3 x = 3 or x = − 3 x=-3 x = − 3 .
Flashcard 7: 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: b 2 − 4 a c b^2-4ac b 2 − 4 a c . Determines the nature and number of solutions.
Flashcard 8: What is the factorization of x 2 − 10 x + 25 x^2-10x+25 x 2 − 10 x + 25 ? Answer: ( x − 5 ) 2 (x-5)^2 ( x − 5 ) 2 . Recognize pattern: ( − 5 ) 2 = 25 (-5)^2 = 25 ( − 5 ) 2 = 25 and 2 ( − 5 ) = − 10 2(-5) = -10 2 ( − 5 ) = − 10 .
Flashcard 9: Solve x 3 − 4 x 2 + 4 x = 0 x^3-4x^2+4x=0 x 3 − 4 x 2 + 4 x = 0 . Answer: x = 0 x=0 x = 0 or x = 2 x=2 x = 2 . Factor out x x x : x ( x 2 − 4 x + 4 ) = x ( x − 2 ) 2 = 0 x(x^2-4x+4)=x(x-2)^2=0 x ( x 2 − 4 x + 4 ) = x ( x − 2 ) 2 = 0 .
Flashcard 10: What is the zero-product property used to solve ( x − 3 ) ( x + 5 ) = 0 (x-3)(x+5)=0 ( x − 3 ) ( x + 5 ) = 0 ? Answer: If a b = 0 ab=0 ab = 0 , then a = 0 a=0 a = 0 or b = 0 b=0 b = 0 . A product equals zero only if at least one factor is zero.
Flashcard 11: What is the discriminant of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , and what does it determine? Answer: b 2 − 4 a c b^2-4ac b 2 − 4 a c ; it determines the number of real solutions. Positive discriminant means 2 real solutions, zero means 1, negative means 0.
Flashcard 12: What is the discriminant for a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , and what does it determine? Answer: Discriminant b 2 − 4 a c b^2-4ac b 2 − 4 a c ; it determines the number of real solutions. Positive means 2 real roots, zero means 1, negative means 0.
Flashcard 13: What are the solutions of ( x − 3 ) ( x + 5 ) = 0 (x-3)(x+5)=0 ( x − 3 ) ( x + 5 ) = 0 ? Answer: x = 3 x=3 x = 3 or x = − 5 x=-5 x = − 5 . Apply zero-product property: set each factor to zero.
Flashcard 14: Identify the solutions of ( x − 3 ) ( x + 5 ) = 0 (x-3)(x+5)=0 ( x − 3 ) ( x + 5 ) = 0 . Answer: x = 3 x=3 x = 3 or x = − 5 x=-5 x = − 5 . Set each factor to zero: x − 3 = 0 x-3=0 x − 3 = 0 gives x = 3 x=3 x = 3 , x + 5 = 0 x+5=0 x + 5 = 0 gives x = − 5 x=-5 x = − 5 .
Flashcard 15: What is the factorization pattern for a perfect square trinomial a 2 + 2 a b + b 2 a^2+2ab+b^2 a 2 + 2 ab + b 2 ? Answer: 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 . Perfect square trinomial factors to squared binomial.
Flashcard 16: Identify the number of real solutions to x 2 − 6 x + 9 = 0 x^2-6x+9=0 x 2 − 6 x + 9 = 0 . Answer: 1 1 1 real solution. ( x − 3 ) 2 = 0 (x-3)^2=0 ( x − 3 ) 2 = 0 has repeated root.
Flashcard 17: What is the maximum number of real solutions a degree n n n polynomial equation can have? Answer: At most n n n real solutions. By the Fundamental Theorem of Algebra, counting multiplicities.
Flashcard 18: Factor completely: x 2 − 5 x − 6 x^2-5x-6 x 2 − 5 x − 6 . Answer: ( x − 6 ) ( x + 1 ) (x-6)(x+1) ( x − 6 ) ( x + 1 ) . Find two numbers that multiply to − 6 -6 − 6 and add to − 5 -5 − 5 .
Flashcard 19: What is the Factor Theorem stated using f ( r ) f(r) f ( r ) and ( x − r ) (x-r) ( x − r ) ? Answer: f ( r ) = 0 f(r)=0 f ( r ) = 0 if and only if ( x − r ) (x-r) ( x − r ) is a factor of f ( x ) f(x) f ( x ) . Root r r r means ( x − r ) (x-r) ( x − r ) divides f ( x ) f(x) f ( x ) .
Flashcard 20: What is the factorization pattern for a perfect square trinomial a 2 − 2 a b + b 2 a^2-2ab+b^2 a 2 − 2 ab + b 2 ? Answer: 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 . Perfect square with subtraction factors to squared difference.
Flashcard 21: How many real solutions does x 2 + 4 x + 5 = 0 x^2+4x+5=0 x 2 + 4 x + 5 = 0 have? Answer: 0 0 0 real solutions. Discriminant = 16 − 20 = − 4 < 0 =16-20=-4<0 = 16 − 20 = − 4 < 0 , so no real solutions exist.
Flashcard 22: What is the standard form of a polynomial equation in x x x of degree n n n ? Answer: a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = 0 a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0 a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = 0 , a n ≠ 0 a_n\neq 0 a n = 0 . Standard form has terms in descending degree order with leading coefficient non-zero.
Flashcard 23: What is the degree of the polynomial 7 x 4 − 3 x 2 + x − 9 7x^4-3x^2+x-9 7 x 4 − 3 x 2 + x − 9 ? Answer: Degree 4 4 4 . The degree is the highest exponent of the variable.
Flashcard 24: Solve the equation ( x − 3 ) ( x + 5 ) = 0 (x-3)(x+5)=0 ( x − 3 ) ( x + 5 ) = 0 . Answer: x = 3 x=3 x = 3 or x = − 5 x=-5 x = − 5 . Apply zero-product property: set each factor equal to zero.
Flashcard 25: Identify the remainder when p ( x ) = x 3 − 4 x + 1 p(x)=x^3-4x+1 p ( x ) = x 3 − 4 x + 1 is divided by ( x − 2 ) (x-2) ( x − 2 ) . Answer: 1 1 1 . By Remainder Theorem: p ( 2 ) = 2 3 − 4 ( 2 ) + 1 = 8 − 8 + 1 = 1 p(2)=2^3-4(2)+1=8-8+1=1 p ( 2 ) = 2 3 − 4 ( 2 ) + 1 = 8 − 8 + 1 = 1 .
Flashcard 26: Identify the solutions of x 2 = 49 x^2=49 x 2 = 49 . Answer: x = 7 x=7 x = 7 and x = − 7 x=-7 x = − 7 . Take square root of both sides.
Flashcard 27: How many real solutions does x 2 + 4 x + 10 = 0 x^2+4x+10=0 x 2 + 4 x + 10 = 0 have? Answer: 0 0 0 real solutions. Discriminant = 16 − 40 = − 24 < 0 =16-40=-24<0 = 16 − 40 = − 24 < 0 , so no real solutions.
Flashcard 28: What is the leading coefficient of − 5 x 3 + 2 x − 1 -5x^3+2x-1 − 5 x 3 + 2 x − 1 ? Answer: − 5 -5 − 5 . The leading coefficient is the coefficient of the highest degree term.
Flashcard 29: What is the relationship between a factor and a root (Factor Theorem)? Answer: ( x − r ) (x-r) ( x − r ) is a factor iff f ( r ) = 0 f(r)=0 f ( r ) = 0 . A polynomial has factor ( x − r ) (x-r) ( x − r ) exactly when r r r is a root.
Flashcard 30: What is the factored form of the difference of squares a 2 − b 2 a^2-b^2 a 2 − b 2 ? Answer: ( a − b ) ( a + b ) (a-b)(a+b) ( a − b ) ( a + b ) . This is the difference of squares factorization pattern.
Flashcard 31: If x = 3 x=3 x = 3 is a solution of P ( x ) = 0 P(x)=0 P ( x ) = 0 , what factor must P ( x ) P(x) P ( x ) have? Answer: Factor ( x − 3 ) (x-3) ( x − 3 ) . By the Factor Theorem, if P ( 3 ) = 0 P(3)=0 P ( 3 ) = 0 , then ( x − 3 ) (x-3) ( x − 3 ) divides P ( x ) P(x) P ( x ) .
Flashcard 32: What does the Remainder Theorem say is the remainder when dividing f ( x ) f(x) f ( x ) by ( x − c ) (x-c) ( x − c ) ? Answer: Remainder = f ( c ) =f(c) = f ( c ) . When dividing f ( x ) f(x) f ( x ) by ( x − c ) (x-c) ( x − c ) , the remainder equals f ( c ) f(c) f ( c ) .
Flashcard 33: Solve for x x x : x 3 − 4 x 2 = 0 x^3-4x^2=0 x 3 − 4 x 2 = 0 . Answer: x = 0 x=0 x = 0 or x = 4 x=4 x = 4 . Factor out x 2 x^2 x 2 : x 2 ( x − 4 ) = 0 x^2(x-4)=0 x 2 ( x − 4 ) = 0 , so x = 0 x=0 x = 0 (double root) or x = 4 x=4 x = 4 .
Flashcard 34: What is the factored form of the perfect square trinomial a 2 − 2 a b + b 2 a^2-2ab+b^2 a 2 − 2 ab + b 2 ? Answer: ( a − b ) 2 (a-b)^2 ( a − b ) 2 . Perfect square with negative middle term factors as ( a − b ) (a-b) ( a − b ) squared.
Flashcard 35: What is the zero-product property used to solve factored equations? Answer: If a b = 0 ab=0 ab = 0 , then a = 0 a=0 a = 0 or b = 0 b=0 b = 0 . At least one factor must equal zero for the product to be zero.
Flashcard 36: Identify the solutions of ( x − 4 ) ( x + 1 ) = 0 (x-4)(x+1)=0 ( x − 4 ) ( x + 1 ) = 0 . Answer: x = 4 x=4 x = 4 and x = − 1 x=-1 x = − 1 . Set each factor to zero: x − 4 = 0 x-4=0 x − 4 = 0 gives x = 4 x=4 x = 4 , x + 1 = 0 x+1=0 x + 1 = 0 gives x = − 1 x=-1 x = − 1 .
Flashcard 37: What is the standard form of a polynomial in descending powers of x x x ? Answer: a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 . Terms arranged from highest to lowest degree.
Flashcard 38: Solve x 2 − 5 x + 6 = 0 x^2-5x+6=0 x 2 − 5 x + 6 = 0 . Answer: x = 2 x=2 x = 2 or x = 3 x=3 x = 3 . Factor as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3)=0 ( x − 2 ) ( x − 3 ) = 0 , then apply zero-product property.
Flashcard 39: Identify the solutions of x 2 + 2 x − 3 = 0 x^2+2x-3=0 x 2 + 2 x − 3 = 0 . Answer: x = 1 x=1 x = 1 and x = − 3 x=-3 x = − 3 . Factor as ( x − 1 ) ( x + 3 ) = 0 (x-1)(x+3)=0 ( x − 1 ) ( x + 3 ) = 0 .
Flashcard 40: What is the product of the roots of 3 x 2 + 6 x − 9 = 0 3x^2+6x-9=0 3 x 2 + 6 x − 9 = 0 ? Answer: c a = − 9 3 = − 3 \frac{c}{a}=\frac{-9}{3}=-3 a c = 3 − 9 = − 3 . For a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , product of roots equals c a \frac{c}{a} a c .
Flashcard 41: What is the degree of the polynomial 7 x 5 − 3 x 2 + 9 7x^5-3x^2+9 7 x 5 − 3 x 2 + 9 ? Answer: 5 5 5 . The degree is the highest power of x x x in the polynomial.
Flashcard 42: 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 x=2 x = 2 or x = 3 x=3 x = 3 . Factors to ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3)=0 ( x − 2 ) ( x − 3 ) = 0 , giving these two solutions.
Flashcard 43: 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 . Formula gives solutions to any quadratic equation a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 .
Flashcard 44: What is the definition of a polynomial in x x x ? Answer: An expression a n x n + ⋯ + a 1 x + a 0 a_nx^n+\cdots+a_1x+a_0 a n x n + ⋯ + a 1 x + a 0 with n n n a nonnegative integer. Each term has form a i x i a_ix^i a i x i where i i i ranges from 0 to n n n .
Flashcard 45: Use the Factor Theorem: if P ( 2 ) = 0 P(2)=0 P ( 2 ) = 0 , what factor divides P ( x ) P(x) P ( x ) ? Answer: ( x − 2 ) (x-2) ( x − 2 ) . By Factor Theorem, P ( 2 ) = 0 P(2)=0 P ( 2 ) = 0 means ( x − 2 ) (x-2) ( x − 2 ) is a factor.
Flashcard 46: Identify the zeros of f ( x ) = ( x − 2 ) ( x + 5 ) f(x)=(x-2)(x+5) f ( x ) = ( x − 2 ) ( x + 5 ) . Answer: x = 2 x=2 x = 2 and x = − 5 x=-5 x = − 5 . Set each factor equal to zero: ( x − 2 ) = 0 (x-2)=0 ( x − 2 ) = 0 and ( x + 5 ) = 0 (x+5)=0 ( x + 5 ) = 0 .
Flashcard 47: What does it mean for r r r to be a solution (root) of p ( x ) = 0 p(x)=0 p ( x ) = 0 ? Answer: p ( r ) = 0 p(r)=0 p ( r ) = 0 . When r r r is substituted for x x x , the polynomial equals zero.
Flashcard 48: 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: 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 49: Find the sum of the solutions of x 2 − 7 x + 10 = 0 x^2-7x+10=0 x 2 − 7 x + 10 = 0 without solving fully. Answer: 7 7 7 . By Vieta's formulas, sum of roots equals negative coefficient of x x x divided by leading coefficient.
Flashcard 50: How many real solutions does x 2 + 4 x + 8 = 0 x^2+4x+8=0 x 2 + 4 x + 8 = 0 have? Answer: 0 0 0 real solutions. Discriminant = 16 − 32 = − 16 < 0 =16-32=-16<0 = 16 − 32 = − 16 < 0 , so no real solutions.
Flashcard 51: Solve x 2 − 5 x + 6 = 0 x^2-5x+6=0 x 2 − 5 x + 6 = 0 . Answer: x = 2 x=2 x = 2 or x = 3 x=3 x = 3 . Factor as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3)=0 ( x − 2 ) ( x − 3 ) = 0 , then solve each factor.
Flashcard 52: What is the factored form of x 2 + 6 x + 9 x^2+6x+9 x 2 + 6 x + 9 ? Answer: ( x + 3 ) 2 (x+3)^2 ( x + 3 ) 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 53: Solve for x x x : x 3 − 4 x = 0 x^3-4x=0 x 3 − 4 x = 0 . Answer: x = − 2 x=-2 x = − 2 , x = 0 x=0 x = 0 , or x = 2 x=2 x = 2 . Factor as x ( x 2 − 4 ) = x ( x − 2 ) ( x + 2 ) = 0 x(x^2-4)=x(x-2)(x+2)=0 x ( x 2 − 4 ) = x ( x − 2 ) ( x + 2 ) = 0 .
Flashcard 54: State the zero-product property used to solve factored equations. Answer: If a b = 0 ab=0 ab = 0 , then a = 0 a=0 a = 0 or b = 0 b=0 b = 0 . A product equals zero only if at least one factor is zero.
Flashcard 55: What is the factored form of x 2 + 6 x + 9 x^2+6x+9 x 2 + 6 x + 9 ? Answer: ( x + 3 ) 2 (x+3)^2 ( x + 3 ) 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 56: What is the factored form of the perfect square trinomial a 2 + 2 a b + b 2 a^2+2ab+b^2 a 2 + 2 ab + b 2 ? Answer: ( a + b ) 2 (a+b)^2 ( a + b ) 2 . Perfect square trinomial: first term squared plus twice the product plus last squared.
Flashcard 57: What is P ( 2 ) P(2) P ( 2 ) for P ( x ) = x 3 − 4 x P(x)=x^3-4x P ( x ) = x 3 − 4 x ? Answer: 0 0 0 . Substitute: P ( 2 ) = 2 3 − 4 ( 2 ) = 8 − 8 = 0 P(2) = 2^3 - 4(2) = 8 - 8 = 0 P ( 2 ) = 2 3 − 4 ( 2 ) = 8 − 8 = 0 .
Flashcard 58: What does a discriminant b 2 − 4 a c < 0 b^2-4ac<0 b 2 − 4 a c < 0 imply about real solutions? Answer: No real solutions. Negative discriminant means no real square roots exist.
Flashcard 59: What is the quadratic formula for 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 . Solves any quadratic equation when factoring is difficult.
Flashcard 60: What is f ( 2 ) f(2) f ( 2 ) for f ( x ) = x 3 − 4 x 2 + x + 6 f(x)=x^3-4x^2+x+6 f ( x ) = x 3 − 4 x 2 + x + 6 ? Answer: 0 0 0 . f ( 2 ) = 8 − 16 + 2 + 6 = 0 f(2) = 8-16+2+6 = 0 f ( 2 ) = 8 − 16 + 2 + 6 = 0 , confirming ( x − 2 ) (x-2) ( x − 2 ) is a factor.
Flashcard 61: What are the solutions of x 2 = 49 x^2=49 x 2 = 49 ? Answer: x = 7 x=7 x = 7 or x = − 7 x=-7 x = − 7 . Take square root of both sides: x = ± 49 x=\pm\sqrt{49} x = ± 49 .
Flashcard 62: Factor completely: x 2 − 16 x^2-16 x 2 − 16 . Answer: ( x − 4 ) ( x + 4 ) (x-4)(x+4) ( x − 4 ) ( x + 4 ) . Difference of squares: x 2 − 4 2 = ( x − 4 ) ( x + 4 ) x^2-4^2=(x-4)(x+4) x 2 − 4 2 = ( x − 4 ) ( x + 4 ) .
Flashcard 63: What is the difference of squares factoring pattern? 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 ) . Factors any expression of form a 2 − b 2 a^2-b^2 a 2 − b 2 .
Flashcard 64: What are the solutions of x 2 − 5 x + 6 = 0 x^2-5x+6=0 x 2 − 5 x + 6 = 0 ? Answer: x = 2 x=2 x = 2 or x = 3 x=3 x = 3 . Factor as ( x − 2 ) ( x − 3 ) = 0 (x-2)(x-3)=0 ( x − 2 ) ( x − 3 ) = 0 or use quadratic formula.
Flashcard 65: Which value is a root of f ( x ) = x 3 − 4 x 2 − x + 4 f(x)=x^3-4x^2-x+4 f ( x ) = x 3 − 4 x 2 − x + 4 : x = 1 x=1 x = 1 or x = 2 x=2 x = 2 ? Answer: x = 1 x=1 x = 1 . f ( 1 ) = 1 − 4 − 1 + 4 = 0 f(1)=1-4-1+4=0 f ( 1 ) = 1 − 4 − 1 + 4 = 0 , so x = 1 x=1 x = 1 is a root.
Flashcard 66: State the quadratic formula for solving 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 . Solves a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 when factoring is difficult.
Flashcard 67: What is the definition of a polynomial equation in x x x ? Answer: An equation setting a polynomial in x x x equal to 0 0 0 . A polynomial expression set equal to zero forms a polynomial equation.
Flashcard 68: How many real solutions does x 2 + 4 x + 5 = 0 x^2+4x+5=0 x 2 + 4 x + 5 = 0 have? Answer: 0 0 0 . Discriminant = 16 − 20 = − 4 < 0 = 16-20 = -4 < 0 = 16 − 20 = − 4 < 0 , so no real solutions exist.
Flashcard 69: What is the leading coefficient of − 4 x 3 + 2 x − 1 -4x^3+2x-1 − 4 x 3 + 2 x − 1 ? Answer: − 4 -4 − 4 . The leading coefficient is the coefficient of the highest degree term.
Flashcard 70: Factor the trinomial x 2 + 7 x + 12 x^2+7x+12 x 2 + 7 x + 12 . Answer: ( x + 3 ) ( x + 4 ) (x+3)(x+4) ( x + 3 ) ( x + 4 ) . Find two numbers that multiply to 12 12 12 and add to 7 7 7 : 3 3 3 and 4 4 4 .
Flashcard 71: What is the degree of the polynomial 7 x 4 − 3 x 2 + x − 9 7x^4-3x^2+x-9 7 x 4 − 3 x 2 + x − 9 ? Answer: 4 4 4 . The degree is the highest exponent of the variable.
Flashcard 72: Solve x 2 − 9 = 0 x^2-9=0 x 2 − 9 = 0 . Answer: x = 3 x=3 x = 3 or x = − 3 x=-3 x = − 3 . Factor as ( x − 3 ) ( x + 3 ) = 0 (x-3)(x+3)=0 ( x − 3 ) ( x + 3 ) = 0 , then apply zero product property.
Flashcard 73: What is the result of factoring x 2 + 7 x + 12 x^2+7x+12 x 2 + 7 x + 12 completely? Answer: ( x + 3 ) ( x + 4 ) (x+3)(x+4) ( x + 3 ) ( x + 4 ) . Find two numbers that multiply to 12 12 12 and add to 7 7 7 : 3 3 3 and 4 4 4 .
Flashcard 74: What is the constant term of 3 x 4 − x + 8 3x^4-x+8 3 x 4 − x + 8 ? Answer: 8 8 8 . The constant term is the term without any variable.
Flashcard 75: What is the maximum possible number of real solutions to a degree n n n polynomial equation? Answer: At most n n n real solutions. Fundamental Theorem: degree n n n polynomial has at most n n n roots.
Flashcard 76: Solve the polynomial equation ( x − 4 ) ( x + 1 ) = 0 (x-4)(x+1)=0 ( x − 4 ) ( x + 1 ) = 0 . Answer: x = 4 x=4 x = 4 or x = − 1 x=-1 x = − 1 . Set each factor to zero: x − 4 = 0 x-4=0 x − 4 = 0 or x + 1 = 0 x+1=0 x + 1 = 0 .
Flashcard 77: What is the leading coefficient of − 5 x 3 + 2 x − 1 -5x^3+2x-1 − 5 x 3 + 2 x − 1 ? Answer: − 5 -5 − 5 . The leading coefficient is the coefficient of the highest degree term.
Flashcard 78: Solve for x x x : x 2 − 6 x + 5 = 0 x^2-6x+5=0 x 2 − 6 x + 5 = 0 . Answer: x = 1 x=1 x = 1 or x = 5 x=5 x = 5 . Factor as ( x − 1 ) ( x − 5 ) = 0 (x-1)(x-5)=0 ( x − 1 ) ( x − 5 ) = 0 , then apply zero-product property.
Flashcard 79: What is the factorization pattern for a difference of squares a 2 − b 2 a^2-b^2 a 2 − b 2 ? 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 ) . Difference of squares factors into sum times difference.
Flashcard 80: What is the Remainder Theorem for dividing f ( x ) f(x) f ( x ) by ( x − r ) (x-r) ( x − r ) ? Answer: The remainder equals f ( r ) f(r) f ( r ) . Substitute r r r into f ( x ) f(x) f ( x ) to get remainder.
Flashcard 81: How many real solutions does x 2 + 4 x + 5 = 0 x^2+4x+5=0 x 2 + 4 x + 5 = 0 have? Answer: 0 0 0 real solutions. Discriminant = 16 − 20 = − 4 < 0 =16-20=-4<0 = 16 − 20 = − 4 < 0 , so no real solutions.
Flashcard 82: What is the perfect square trinomial pattern for a 2 + 2 a b + b 2 a^2+2ab+b^2 a 2 + 2 ab + b 2 ? Answer: ( a + b ) 2 (a+b)^2 ( a + b ) 2 . Recognizes when a trinomial is a perfect square.
Flashcard 83: Solve for x x x : x 2 = 16 x^2=16 x 2 = 16 . Answer: x = ± 4 x=\pm^4 x = ± 4 . Take square root of both sides: x = ± 16 = ± 4 x=\pm\sqrt{16}=\pm^4 x = ± 16 = ± 4 .
Flashcard 84: State the quadratic formula for solving 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 . Formula solves any quadratic equation.
Flashcard 85: What is the factored form of the difference of squares a 2 − b 2 a^2-b^2 a 2 − b 2 ? Answer: ( a − b ) ( a + b ) (a-b)(a+b) ( a − b ) ( a + b ) . Difference of squares factors as product of sum and difference.
Flashcard 86: What is the quadratic formula for solutions of 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 . General formula for solving any quadratic equation.
Flashcard 87: What is the discriminant of a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , and what does it determine? Answer: Discriminant b 2 − 4 a c b^2-4ac b 2 − 4 a c ; it determines the number of real solutions. Positive gives 2 real roots, zero gives 1, negative gives 0.
Flashcard 88: What is the sum of the solutions of x 2 − 6 x + 11 = 0 x^2-6x+11=0 x 2 − 6 x + 11 = 0 using − b a -\frac{b}{a} − a b ? Answer: 6 6 6 . For a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , sum of roots equals − b a = − − 6 1 = 6 -\frac{b}{a} = -\frac{-6}{1} = 6 − a b = − 1 − 6 = 6 .
Flashcard 89: What is the discriminant of 2 x 2 − 3 x + 5 = 0 2x^2-3x+5=0 2 x 2 − 3 x + 5 = 0 ? Answer: − 31 -31 − 31 . Calculate b 2 − 4 a c = ( − 3 ) 2 − 4 ( 2 ) ( 5 ) = 9 − 40 b^2-4ac = (-3)^2-4(2)(5) = 9-40 b 2 − 4 a c = ( − 3 ) 2 − 4 ( 2 ) ( 5 ) = 9 − 40 .
Flashcard 90: What does the Zero Product Property state for a b = 0 ab=0 ab = 0 ? Answer: If a b = 0 ab=0 ab = 0 , then a = 0 a=0 a = 0 or b = 0 b=0 b = 0 . At least one factor must be zero.
Flashcard 91: State the zero-product property for a b = 0 ab=0 ab = 0 . Answer: If a b = 0 ab=0 ab = 0 , then a = 0 a=0 a = 0 or b = 0 b=0 b = 0 . A product equals zero only if at least one factor equals zero.
Flashcard 92: Identify the solutions of x 2 − 5 x + 6 = 0 x^2-5x+6=0 x 2 − 5 x + 6 = 0 . Answer: 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 .
Flashcard 93: What is the complete factorization of x 2 − 49 x^2-49 x 2 − 49 over the integers? Answer: ( x − 7 ) ( x + 7 ) (x-7)(x+7) ( x − 7 ) ( x + 7 ) . 49 = 7 2 49 = 7^2 49 = 7 2 , so apply difference of squares pattern.
Flashcard 94: Factor completely: x 2 + 6 x + 9 x^2+6x+9 x 2 + 6 x + 9 . Answer: ( x + 3 ) 2 (x+3)^2 ( x + 3 ) 2 . Recognizes as ( x ) 2 + 2 ( x ) ( 3 ) + ( 3 ) 2 (x)^2+2(x)(3)+(3)^2 ( x ) 2 + 2 ( x ) ( 3 ) + ( 3 ) 2 , a perfect square trinomial.
Flashcard 95: What is the degree of the polynomial 7 x 4 − 3 x 2 + 9 7x^4-3x^2+9 7 x 4 − 3 x 2 + 9 ? Answer: 4 4 4 . The degree is the highest power of the variable.
Flashcard 96: What is the product of the solutions of 2 x 2 + 3 x − 5 = 0 2x^2+3x-5=0 2 x 2 + 3 x − 5 = 0 using c a \frac{c}{a} a c ? Answer: − 5 2 -\frac{5}{2} − 2 5 . For a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 , product of roots equals c a = − 5 2 \frac{c}{a} = \frac{-5}{2} a c = 2 − 5 .
Flashcard 97: What does a discriminant value b 2 − 4 a c < 0 b^2-4ac<0 b 2 − 4 a c < 0 imply about real solutions? Answer: No real solutions. Negative discriminant means no real roots exist.
Flashcard 98: Identify the number of real solutions to x 2 + 4 x + 5 = 0 x^2+4x+5=0 x 2 + 4 x + 5 = 0 . Answer: 0 0 0 real solutions. Discriminant 16 − 20 = − 4 < 0 16-20=-4<0 16 − 20 = − 4 < 0 .
Flashcard 99: 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: b 2 − 4 a c b^2-4ac b 2 − 4 a c . Determines the nature and number of roots for quadratics.
Flashcard 100: State the quadratic formula for solutions of 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 . Solves any quadratic equation when a ≠ 0 a\neq 0 a = 0 .