Question 1·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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The cubic function $p$ is defined by $p(x)=ax^3+bx^2+cx+d$ , where $a$ , $b$ , $c$ , and $d$ are constants and $a\ne0$ . The graph of $y=p(x)$ has $x$ -intercepts at $(-4,0)$ and $(2,0)$ , and $p(1)=p(-1)$ .

Which choice gives the value of $b$ in terms of $a$ ?

Known roots, inputs where a polynomial equals $0$ , give factors of a cubic. Write an unknown third factor, then use the equal function outputs to find its constant with a Desmos regression. Two roots supply two factors; the output condition fixes the third. Don't mistake that third constant for the requested coefficient: all three factors contribute to it.

Hints

Hint 1
Each root $r$ gives a factor $x-r$ . The intercepts at $-4$ and $2$ give $x+4$ and $x-2$ . What third linear factor would let their product remain a cubic?
Hint 2
Plug $1$ and $-1$ into the factored expression. The problem says those outputs are equal, so set them equal and cancel $a$ , which you can do because $a\ne0$ .
Hint 3
The unknown in the third factor isn't necessarily $b$ . In the original cubic, $b$ multiplies $x^2$ . Which products from all three factors create an $x^2$ term?

Step-by-step

Approach 1: Build the cubic from its roots

Step 1
Write a factor for each known root
An $x$ -intercept has output $0$ , so the roots $-4$ and $2$ give the factors $x+4$ and $x-2$ . A cubic needs one more linear factor. Call it $x+t$ , and put $a$ in front to keep the given leading coefficient: $p(x)=a(x+4)(x-2)(x+t)$ . The third factor is what the remaining condition will determine.
Step 2
Evaluate the two equal outputs
Substitute $1$ into the three factors: $p(1)=a(1+4)(1-2)(1+t)$ . Multiply the numbers: $p(1)=-5a(1+t)$ . Now substitute $-1$ : $p(-1)=a(-1+4)(-1-2)(-1+t)$ . Multiply the numbers: $p(-1)=-9a(t-1)$ . Rewrite $t-1$ as $-(1-t)$ : $p(-1)=9a(1-t)$ .
Step 3
Turn equal outputs into an equation
The outputs are equal, so set the expressions you found equal: $-5a(1+t)=9a(1-t)$ . Divide by $a$ , which is allowed because the problem says $a\ne0$ : $-5(1+t)=9(1-t)$ .
Step 4
Find the third factor's constant
Type $-5(1+t)\sim9(1-t)$ in Desmos. The $\sim$ tells Desmos to find the unknown $t$ that makes the sides equal. Under PARAMETERS, it shows $t=3.5$ . That's the constant in the third factor, not yet $b/a$ .
Step 5
Find the coefficient of $x^2$
Multiply the first two factors: $(x+4)(x-2)=x^2-2x+4x-8$ . Combine the $x$ terms: $(x+4)(x-2)=x^2+2x-8$ . When you multiply by $x+t$ , only these products make $x^2$ : $t(x^2)+x(2x)=(t+2)x^2$ . So $b=a(t+2)$ . Add $t+2$ in Desmos; it prints $5.5$ , or $\frac{11}{2}$ using the fraction button. The requested coefficient is $b=\frac{11}{2}a$ . Choice C.

Approach 2: Cancel coefficients using the given conditions

Step 1
Use the equal outputs to find $c$
At $1$ and $-1$ , the $bx^2$ and $d$ terms stay the same, so they cancel when you subtract the outputs. Subtract $p(-1)$ from $p(1)$ : $p(1)-p(-1)=2a+2c$ . Because those outputs are equal, set the difference to $0$ : $2a+2c=0$ . Divide by $2$ : $a+c=0$ . Subtract $a$ : $c=-a$ .
Step 2
Write an equation for each intercept
Each intercept has output $0$ . Substitute the two given inputs into the cubic: $p(-4)=-64a+16b-4c+d=0$ , $p(2)=8a+4b+2c+d=0$ . Replace $c$ with $-a$ in both equations: $-60a+16b+d=0$ , $6a+4b+d=0$ .
Step 3
Eliminate $d$ and solve for $b$
Subtract the second equation from the first to remove $d$ : $(-60a+16b+d)-(6a+4b+d)=0$ . Combine like terms: $-66a+12b=0$ . Add $66a$ : $12b=66a$ . Divide by $12$ : $b=\frac{11}{2}a$ . That's the coefficient the question asks for. Choice C.

Question 1·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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