To square a binomial on the ACT Math test, multiply the expression by itself using the FOIL method and then combine like terms. This is a core skill in ACT Math polynomials, especially in questions involving expansion and simplification.A common mistake students make is squaring each term separately instead of properly expanding the binomial. Avoiding this error is essential for improving accuracy and saving time on the test.Another effective strategy is substituting values for variables to quickly verify answers or simplify expressions when needed.
For example, expanding $(3x+8y^2)^2$ yields: $$ (3x+8y^2)^2 = 9x^2+48xy^2+64y^4. $$
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Solving higher-degree polynomial equations typically involves factoring to reduce the degree of the expression. Once the polynomial is reduced to a quadratic form, apply the quadratic formula or further factoring. If one solution is already known, use polynomial long division or synthetic division to find the remaining roots.
For example, to solve $x^4-13x^2+36=0$, treat the equation as a quadratic in terms of $x^2$. Factoring the expression yields: $$ (x^2-9)(x^2-4) = 0. $$ Setting each factor to zero results in $x^2=9$ and $x^2=4$, which gives the solution set $\{-3, -2, 2, 3\}$.
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When working with polynomials, a fundamental tool is the difference of squares formula: $$ a^2-b^2=(a+b)(a-b). $$ In higher-degree polynomials, this formula may require repeated application. The concept also extends to expressions involving irrational conjugates or complex numbers. Examples include: \begin{align*} 9x^2-25 & = (3x+5)(3x-5), \\ x^2-\dfrac{25}{49} & = \left(x+\dfrac{5}{7}\right)\left(x-\dfrac{5}{7}\right), \\ x^2-5 & = x^2-(\sqrt{5})^2 = (x+\sqrt{5})(x-\sqrt{5}), \\ x^2+4 & = x^2-(-4)=(x+2i)(x-2i). \end{align*} Additionally, familiarity with the sum and difference of cubes formulas is advantageous, although they may be provided or the problem might be solvable using other methods: \begin{align*} a^3+b^3 & =(a+b)(a^2-ab+b^2) \\ a^3-b^3 &= (a-b)(a^2+ab+b^2) \end{align*} For example, factoring $8x^3+27$ yields $(2x+3)(4x^2-6x+9)$. These identities can be verified by FOILing the right side.
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To construct a polynomial from its roots, multiply the linear factors corresponding to each solution. For instance, a fifth-degree polynomial with roots $2, 4, 11, 23,$ and $32$ is expressed as: $$ (x-2)(x-4)(x-11)(x-23)(x-32). $$ Consider finding a third-degree polynomial with integer coefficients that has solutions $3$ and $5+\sqrt{2}$. Since irrational roots occur in conjugate pairs, $5-\sqrt{2}$ must also be a solution. Form the factors by subtracting each root from $x$: \begin{align*} & \Big(x-\big(5+\sqrt{2}\big)\Big) \Big(x-\big(5-\sqrt{2}\big)\Big)(x-3) \\ & = \big((x-5)-\sqrt{2}\big)\big((x-5)+\sqrt{2}\big)(x-3) \end{align*} Expand the conjugate factors using the difference of squares formula: \begin{align*} & \big((x-5)^2 - (\sqrt{2})^2\big) (x-3) \\ & = (x^2-10x+25-2)(x-3) \\ & = (x^2-10x+23)(x-3) \end{align*} Finally, expand the remaining terms to find the polynomial: \begin{align*} & \ x^3-3x^2-10x^2+30x+23x-69 \\ & = x^3-13x^2+53x-69. \end{align*}
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As illustrated in the figures, the behavior of a graph at an intercept reveals the multiplicity of the root. If the graph intersects the $x$-axis directly without turning or flattening, there is a single root at that value. If the graph touches the $x$-axis and turns (is tangent), this indicates a double root. If the graph flattens as it crosses the $x$-axis, it indicates a triple root.
The degree of the polynomial corresponds to the exponent of the leading term in the expanded form. For polynomials of even degree, the end behavior is the same on both sides; for odd degree, the end behavior differs. Additionally, if the leading coefficient is negative, the standard end behavior is reversed.
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The Factor Theorem can be used to determine unknown coefficients in a polynomial. For example, suppose that $x+5$ is a factor of the cubic polynomial $x^3+2x^2+3x+k$. To find the value of the constant $k$, substitute the root $x=-5$ into the expression and set it to zero: \begin{align*} & (-5)^3+2(-5)^2+3(-5) +k=0 \\ \implies & -125+50-15+k=0 \\ \implies & -90+k=0 \\ \implies & k=90. \end{align*}
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To find the coefficients in a binomial expansion, apply the Binomial Theorem: $$ (x+y)^n = \sum_{k=0}^n {}_{n}\mathrm{C}_{k} x^{n-k}y^k. $$ Substituting the values for $(x+5)^4$ yields: \begin{align*} &(x+5)^4 \\ = \ & {}_{4}\mathrm{C}_{0}x^4+ {}_{4}\mathrm{C}_{1}x^3\cdot5+ {}_{4}\mathrm{C}_{2}x^2\cdot5^2+ {}_{4}\mathrm{C}_{3}x\cdot5^3 \\ & + {}_{4}\mathrm{C}_{4}x^0\cdot5^4 \\ = \ & 1\cdot x^4+4\cdot x^3\cdot5+ 6\cdot x^2\cdot5^2+4\cdot x\cdot5^3+5^4 \\ = \ & x^4+20x^3+150x^2+500x+625. \end{align*} Mastery of the Binomial Theorem is advantageous for high-scoring students. However, exam questions likely involve exponents small enough to permit manual expansion.
Certain problems may not require the full theorem, such as finding the coefficient of the $x^5$ term in $(2x + 7)^5$. Since this corresponds to the leading term, simply raise the variable term to the given power: $$ (2x)^5 = 32x^5 $$ The coefficient is 32.
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Answer key: 1.D, 2.B, 3.A, 4.C, 5.C, 6.B, 7.D, 8.B, 9.A, 10.A, 11.B, 12.B, 13.D, 14.B, 15.D, 16.C, 17.D, 18.C, 19.B, 20.B,
