. Given that , , , and are integers, which choice must be true?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 151 | Free SAT Prep | Aniko

00:00

Question 151·200 Super-Hard SAT Math Questions·Advanced Math

0 / 200

$8x^2+bx+15=(ax-m)(cx+l)$ . Given that $a$ , $c$ , $m$ , and $l$ are integers, which choice must be true?

For “equivalent expressions” questions with a factored form, expand the product and match coefficients. When the question asks what must be true, translate the constraints (like $ac=8$ and $ml=-15$ ) into a small set of cases (factor pairs), then use parity/modular reasoning (often mod 2 or mod 4) to find an invariant that holds in every case.

Hints

Click me!

Expand and compare

Expand $(ax-m)(cx+l)$ and match the $x^2$ , $x$ , and constant terms with $8x^2+bx+15$ .

Use the constant term first

From the constant term you’ll get an equation involving $m$ and $l$ . Use it to decide whether $m$ and $l$ must be even or odd.

Consider the possibilities for $ac=8$

List the integer factor pairs of 8 (including sign). Think about whether $a$ and $c$ are odd/even in each case.

Work modulo 4

Using $b=al-mc$ and the parity information, decide whether $b$ can ever be divisible by 4.

Desmos Guide

Create lists of possible integer factor pairs

In Desmos, enter lists for possible values:

A=[-8,-4,-2,-1,1,2,4,8]

Then compute the matching list:

C=8/A

(These are all integers because each value in A divides 8.)

List possible $l$ values and compute $m$

Enter the divisors of 15:

L=[-15,-5,-3,-1,1,3,5,15]

Then compute:

M=-15/L

Test examples with sliders (quick check)

Create sliders for indices i and j (integers) so you can pick one value from each list:

set i from 1 to length(A)
set j from 1 to length(L)

Define:

a=A[i], c=C[i], l=L[j], m=M[j]

Then compute:

b=a*l-m*c

Check divisibility by 4 across cases

Move the sliders through different valid combinations and compute the remainder when dividing by 4 using Desmos syntax, for example:

r=mod(b,4)

You’ll observe r is never 0, which supports that $b$ is not a multiple of 4.

Step-by-step Explanation

Expand and match coefficients

Expand:

\begin{aligned} (ax-m)(cx+l) &= a c x^2 + (al-mc)x - ml. \end{aligned}

Match with $8x^2+bx+15$ to get

$ac=8$
$-ml=15\ \Rightarrow\ ml=-15$
$b=al-mc$

Use the constant term to get parity for $m$ and $l$

Since $ml=-15$ is odd, both $m$ and $l$ must be odd integers.

List the possible parities of $a$ and $c$ from $ac=8$

Because $ac=8$ , the integer pairs $(a,c)$ must make one of these parity situations happen:

One of $a,c$ is odd and the other is even (e.g., $1\cdot 8$ or $8\cdot 1$ , with possible negative signs).
Both $a$ and $c$ are even (e.g., $2\cdot 4$ or $4\cdot 2$ , with possible negative signs).

Determine $b \bmod 4$ in each situation

Recall $b=al-mc$ with $m,l$ odd.

Case 1: one of $a,c$ is odd and the other is even. Then $al$ is odd $\times$ odd $=$ odd, while $mc$ is odd $\times$ even $=$ even, so $b=$ odd $-$ even $=$ odd. Therefore, $b$ cannot be divisible by 4.

Case 2: both $a$ and $c$ are even. The only (absolute) factor pairs are $(2,4)$ or $(4,2)$ .

If $(a,c)=(2,4)$ , then $b=2l-4m=2(l-2m)$ . Here $l$ is odd and $2m$ is even, so $l-2m$ is odd, making $b\equiv 2\pmod 4$ .
If $(a,c)=(4,2)$ , then $b=4l-2m=2(2l-m)$ . Here $2l$ is even and $m$ is odd, so $2l-m$ is odd, again making $b\equiv 2\pmod 4$ .

Conclude the statement that must be true

In Case 1, $b$ is odd, so it is not divisible by 4. In Case 2, $b\equiv 2\pmod 4$ , so it is also not divisible by 4.

Therefore, the statement that must be true is $b$ is not a multiple of 4.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation