Question 57·200 Super-Hard SAT Math Questions·Advanced Math

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20x^{12} + kx^6 + 45

The expression above can be written as $(ax^6 + b)(cx^6 + d)$ , where $a$ , $b$ , $c$ , and $d$ are integers. Which choice is the maximum possible value of $k$ ?

The powers $x^{12}$ and $x^6$ reveal a quadratic in $x^6$ : the larger power is the square of the smaller. Expand the proposed product and match coefficients, then use the fixed product of its two middle-term pieces to bound their sum. Check that integer factors reach the bound; a valid factorization need not give the maximum value.

Hints

Hint 1
Treat $x^6$ as one quantity, so $x^{12}=(x^6)^2$ . When you multiply the two factors, which two cross products contribute to the $x^6$ term?
Hint 2
An identity is an equality that holds for every $x$ . Match terms with the same power: the $x^{12}$ coefficient and the constant fix two products, even though the middle coefficient can change.
Hint 3
The two pieces of the middle coefficient have a fixed product. For positive integers $p$ and $q$ , $(p-1)(q-1)\ge0$ puts a limit on their sum. When could that limit be reached?

Step-by-step

Bound the middle coefficient
Step 1
Expand to locate the middle term
Multiply the proposed factors. The cross products are $ax^6\cdot d$ and $b\cdot cx^6$ , so both contribute to the $x^6$ term:
$(ax^6+b)(cx^6+d)=acx^{12}+(ad+bc)x^6+bd.$
Step 2
Match coefficients
The product must be the same expression as the one given, so terms with matching powers have matching coefficients:
$ac=20,\qquad bd=45,\qquad k=ad+bc.$
The question asks you to make the last sum as large as possible, not merely to find one factorization.
Step 3
Find the fixed product of the cross terms
Because $ac$ and $bd$ are positive, $ad$ and $bc$ have the same sign. Two negative cross terms give a negative $k$ , so a maximum uses positive ones. Set $p=ad$ and $q=bc$ ; these are positive integers. Multiply them:
$pq=(ad)(bc).$
Reorder the factors:
$pq=(ac)(bd).$
Use $ac=20$ and $bd=45$ :
$pq=20\cdot45.$
Step 4
Bound their sum
Since $p$ and $q$ are positive integers, both $p-1$ and $q-1$ are nonnegative. Expand $(p-1)(q-1)\ge0$ :
$pq-p-q+1\ge0.$
Rearrange to bound the sum you need:
$p+q\le pq+1.$
Use $k=p+q$ :
$k\le pq+1.$
Use $pq=20\cdot45$ :
$k\le20\cdot45+1.$
With a fixed positive integer product, the greatest sum occurs when one factor is $1$ .
Step 5
Reach the bound with integer factors
Choose $a=20$ , $c=1$ , $b=1$ , and $d=45$ . Then $ac=20$ , $bd=45$ , and $bc=1$ , so this choice can reach the bound. In Desmos, type $x_1=[1...5]$ , then $20x_1^{12}+kx_1^6+45\sim(20x_1^6+1)(x_1^6+45)$ . The $\sim$ fits $k$ at those inputs; Desmos reports $k=901$ under PARAMETERS. Type $20\cdot45+1$ to see that the bound is also $901$ . The maximum possible value of $k$ is $901$ . Choice D.

Question 57·200 Super-Hard SAT Math Questions·Advanced Math

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