SAT Question 00165291: Equivalent Expressions answer and explanation with Desmos solution

5$.It’sgiventhat＄6x^4+31x^2+35$ can be rewritten as $(3x^2+a)(2x^2+b)$, where $a$ and $b$ are positive integers. The expression $(3x^2+a)(2x^2+b)$ can be rewritten as ＄6 x^{4} + 3 b x^{2} + 2 a x^{2} + a b$,or＄6x^4+(3b+2a)x^2+ab$. Therefore, the expression ＄6 x^{4} + ( 3 b + 2 a ) x^{2} + a b$isequivalentto＄6x^4+31x^2+35$. It follows that ＄3 b + 2 a = 31$and$a b = 35$.Since$a$and$b$arepositiveintegersand$a b = 35$,thepossiblevaluesof$a$and$b$are$1$and$35$,$5$and$7$,$7$and$5$,and＄35$ and $1$, respectively. Of these options, the only pair that satisfies the equation ＄3 b + 2 a = 31$is$a = 5$and$b = 7$,since＄3(7)+2(5)=31$. It’s also given that ＄6 x^{4} + 31 x^{2} + 35$canberewrittenas$( 3 x^{2} + c ) ( 2 x^{2} + d )$,where$c$and$d$arepositivenonintegers.Theexpression$( 3 x^{2} + c ) ( 2 x^{2} + d )$canberewrittenas＄6x^4+3d{x}^2+2c{x}^2+cd$, or ＄6 x^{4} + ( 3 d + 2 c ) x^{2} + c d$.Therefore,theexpression＄6x^4+(3d+2c)x^2+cd$ is equivalent to ＄6 x^{4} + 31 x^{2} + 35$.Itfollowsthat＄3d+2c=31$ and $cd=35$. Since it's given that $c$ and $d$ are positive nonintegers, it follows that $c$ and $d$ are not zero. Thus, dividing both sides of the equation $cd=35$ by $c$ yields $d=\frac{35}{c}$. Substituting $\frac{35}{c}$ for $d$ in the equation ＄3 d + 2 c = 31$yields$3 \left( \frac{35}{c} \right) + 2 c = 31$,or$\frac{105}{c} + 2 c = 31$.Multiplyingbothsidesofthisequationby$c$yields$c \left( \frac{105}{c} + 2 c \right) = c ( 31 )$,or＄105+2c^2=31c$. Subtracting ＄31 c$frombothsidesofthisequationyields＄105+2c^2-31c=0$, or ＄2 c^{2} - 31 c + 105 = 0$.Thisequationcanberewrittenas＄2c^2-10c-21c+105=0$. Since the first two terms of the expression ＄2 c^{2} - 10 c - 21 c + 105$haveacommonfactorof＄2c$ and the last two terms of the expression ＄2 c^{2} - 10 c - 21 c + 105$haveacommonfactorof＄21$, the expression ＄2 c^{2} - 10 c - 21 c + 105$canberewrittenas＄2c(c-5)-21(c-5)$. Since each term of this expression has a common factor of $(c-5)$, the expression ＄2 c ( c - 5 ) - 21 ( c - 5 )$canberewrittenas$( c - 5 ) ( 2 c - 21 )$.Thus,theequation＄2c^2-31c+105=0$ can be rewritten as $(c-5)(2c-21)=0$. It follows that $c=5$ or $c=\frac{21}{2}$. Since $c$ is a positive noninteger, it follows that $c=\frac{21}{2}$, or $c=10.5$. Substituting ＄5$for$a$and$10.5$for$c$intotheexpression$a + c$yields$5 + 10.5$,or＄15.5$. Therefore, the value of $a+c$ is $15.5$.

Alternate approach: The correct answer is $15.5$. It’s given that $6x^4+31x^2+35$ can be rewritten as $\left(3x^2+a\right)\left(2x^2+b\right)$, where $a$ and $b$ are positive integers. The expression $\left(3x^2+a\right)\left(2x^2+b\right)$ can be rewritten as $6x^4+3b{x}^2+2a{x}^2+ab$, or $6x^4+(3b+2a)x^2+ab$. Therefore, the expression $6x^4+\left(3b+2a\right)x^2+ab$ is equivalent to $6x^4+31x^2+35$. It follows that $3b+2a=31$ and $ab=35$. Since it's given that $a$ and $b$ are positive integers, it follows that $a$ and $b$ are not zero. Thus, dividing both sides of the equation $ab=35$ by $a$ yields $b=\frac{35}{a}$. Substituting $\frac{35}{a}$ for $b$ in the equation $3b+2a=31$ yields $3\left(\frac{35}{a}\right)+2a=31$, or $\frac{105}{a}+2a=31$. Multiplying both sides of this equation by $a$ yields $105+2a^2=31a$. Subtracting $31a$ from both sides of this equation yields $2a^2-31a+105=0$. This equation can be rewritten as $2a^2-10a-21a+105=0$. Since the first two terms of the expression $2a^2-10a-21a+105$ have a common factor of $2a$ and the last two terms of the expression $2a^2-10a-21a+105$ have a common factor of $21$, the expression $2a^2-10a-21a+105$ can be rewritten as $2a(a-5)-21(a-5)$. Since each term of this expression has a common factor of $\left(a-5\right)$, the expression $2a(a-5)-21(a-5)$ can be rewritten as $(a-5)(2a-21)$. Thus, the equation $2a^2-31a+105=0$ can be rewritten as $\left(a-5\right)\left(2a-21\right)=0$. It follows that $a=5$ or $a=\frac{21}{2}$. Since $a$ is a positive integer, it follows that $a=5$. It’s also given that $6x^4+31x^2+35$ can be rewritten as $\left(3x^2+c\right)\left(2x^2+d\right)$, where $c$ and $d$ are positive nonintegers. The expression $\left(3x^2+c\right)\left(2x^2+d\right)$ can be rewritten as $6x^4+3d{x}^2+2c{x}^2+cd$, or $6x^4+(3d+2c)x^2+cd$. Therefore, the expression $6x^4+(3d+2c)x^2+cd$ is equivalent to $6x^4+31x^2+35$. It follows that $3d+2c=31$ and $cd=35$. Since it's given that $c$ and $d$ are positive nonintegers, it follows that $c$ and $d$ are not zero. Thus, dividing both sides of the equation $cd=35$ by $c$ yields $d=\frac{35}{c}$. Substituting $\frac{35}{c}$ for $d$ in the equation $3d+2c=31$ yields $3\left(\frac{35}{c}\right)+2c=31$, or $\frac{105}{c}+2c=31$. Multiplying both sides of this equation by $c$ yields $105+2c^2=31c$. Subtracting $31c$ from both sides of this equation yields $2c^2-31c+105=0$. Since the equations $2a^2-31a+105=0$ and $2c^2-31c+105=0$ represent the same equation, it follows that $a$ is equal to $c$. Therefore, $c=5$ or $c=\frac{21}{2}$. Since $c$ is a positive noninteger, it follows that $c=\frac{21}{2}$, or $c=10.5$. Substituting $5$ for $a$ and $10.5$ for $c$ in the expression $a+c$ yields $5+10.5$, or $15.5$. Therefore, the value of $a+c$ is $15.5$. Note that 15.5 and 31/2 are examples of ways to enter a correct answer.