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

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

A polynomial identity gives the same result on both sides for every $x$ , so matching coefficients connects the numbers in its factored and expanded forms. When those numbers must be integers, their possible factor pairs matter. Use a Desmos regression on valid factorizations to test tempting claims, then check every possible case. An example can disprove a claim, but it cannot prove what must happen.

Hints

Hint 1
In a polynomial identity, terms with the same power of $x$ must match. The $x^2$ term comes from multiplying the two terms containing $x$ . What does matching those terms tell you about $a$ and $c$ ?
Hint 2
An odd product of integers requires both factors to be odd. Match the constant terms, then use that fact to work out whether $m$ and $l$ are odd or even.
Hint 3
The remainder modulo $4$ is what is left after dividing by $4$ . If both $a$ and $c$ are even and their product is $8$ , what remainders can the two products making up $b$ have?

Step-by-step

Match coefficients and check integer cases
Step 1
Match the $x^2$ coefficients
The $x^2$ term on the right comes from $(ax)(cx)=acx^2$ . The two forms are equal for every $x$ , so their $x^2$ coefficients match:
$ac=8.$
Step 2
Match the constants
The constant term on the right is $(-m)l$ , so match it with $15$ : $-ml=15$ . Multiply by $-1$ : $ml=-15$ . Since $m$ and $l$ are integers and their product is odd, both are odd.
Step 3
Find the products that make $b$
The two products containing one $x$ are $(ax)l=alx$ and $(-m)(cx)=-cmx$ . Their sum must match $bx$ , so the coefficient of $x$ is $b=al-cm$ . The question is now about what this difference can be when $ac=8$ and $m,l$ are odd.
Step 4
Test a factorization with an odd coefficient
Choose $a=1$ , $c=8$ , $m=-3$ , and $l=5$ . These integers satisfy $ac=8$ and $ml=-15$ . Type $x_1=[1...5]$ for five inputs, then type $8x_1^2+b_1x_1+15\sim(x_1+3)(8x_1+5)$ . The $\sim$ tells Desmos to fit $b_1$ ; under PARAMETERS it shows $b_1=29$ . So $b$ does not have to be even.
Step 5
Test a factorization with an even coefficient
Keep those lines and add a second fit using $a=2$ , $c=4$ , $m=-3$ , and $l=5$ . Type $8x_1^2+b_2x_1+15\sim(2x_1+3)(4x_1+5)$ . Desmos shows $b_2=22$ under PARAMETERS. So $b$ does not have to be odd or a multiple of $8$ . Examples rule out claims, but you still need to prove what always happens.
Step 6
Handle the case with an odd factor of $8$
If one of $a,c$ is odd, the other is even because $ac=8$ . Since $m,l$ are odd, one of $al,cm$ is odd and the other is even. Their difference $b=al-cm$ is odd, so it cannot be divisible by $4$ .
Step 7
Handle the case with two even factors
If both $a$ and $c$ are even, $ac=8$ means their absolute values are $2$ and $4$ . Because $m,l$ are odd, one of $al,cm$ leaves remainder $2$ when divided by $4$ ; the other leaves remainder $0$ . Their difference $b=al-cm$ leaves remainder $2$ , whichever comes first. Together with the odd case, this covers every allowed factor pair. Test examples to reject claims, but check every case to prove a claim must hold. So $b$ is not a multiple of $4$ . Choice C.

Lessons that teach this

SAT Advanced AlgebraAdvancedCoreUse polynomial identities, factors, and unknown coefficients

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

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