Super-Hard SAT Advanced Math Question 4 | Free SAT Prep

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Question 4·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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The expression $12(3x^2-2x-1)^2+38(3x^2-2x-1)+30$ can be rewritten as $(4(3x^2-2x-1)+a)(3(3x^2-2x-1)+b)$ , where $a$ and $b$ are positive integers. It can also be rewritten as $(4(3x^2-2x-1)+c)(3(3x^2-2x-1)+d)$ , where $c$ and $d$ are positive nonintegers. What is the value of $3a+2c$ ?

Your answer

When an expression contains the same complicated quantity several times, rename that quantity with a simple variable. Then use the structure of a product of two binomials: the constant terms multiply to the constant term, and the two cross-products add to the middle coefficient. Solve for the possible values, then use any integer or noninteger conditions to assign the correct value to each variable.

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Simplify the repeated expression

Replace $3x^2-2x-1$ with a single variable, such as $u$ , before comparing the expressions.

Match coefficients

After expanding $(4u+a)(3u+b)$ , compare the coefficient of $u$ and the constant term with those in the original expression.

Use the integer restriction last

Eliminate $b$ using the product condition. The resulting quadratic has one integer solution and one positive noninteger solution for the constant in the first factor.

Desmos Guide

Use algebra first

Algebra is faster because the repeated expression can be replaced by one variable. After matching coefficients, the possible values of the constant in the first factor satisfy $3x^2-38x+120=0$ .

Graph the quadratic

Enter $y=3x^2-38x+120$ . Find the two x-intercepts. These represent the two possible values for $a$ or $c$ .

Apply the restrictions and verify

Choose the integer intercept for $a$ and the positive noninteger intercept for $c$ . Then enter $3a+2c$ as a calculation to verify the result.

Step-by-step Explanation

Use a temporary variable

Let $u=3x^2-2x-1$ . Expanding the first factored form gives $(4u+a)(3u+b)=12u^2+(4b+3a)u+ab$ . Therefore, matching coefficients with $12u^2+38u+30$ gives $ab=30$ and $3a+4b=38$ .

Find the possible values of the first constant

Since $ab=30$ , substitute $b=30/a$ into $3a+4b=38$ . This gives $3a+120/a=38$ , or $3a^2-38a+120=0$ . Factoring gives $(3a-20)(a-6)=0$ , so the possible values in the first factor are $6$ and $20/3$ .

Apply the number-type conditions

Because $a$ is an integer, $a=6$ . The corresponding noninteger value is $c=20/3$ . Then $3a+2c=3(6)+2(20/3)=94/3$ .

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Step-by-step Explanation

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