Question

Let $P\left(x\right)=4x^2+9y^2+16z^2+12xy+24yz+16xz$ factorize $P\left(x\right)$ for $x=1,y=1,z=1.$

• A. $36$
• B. $49$
• C. $64$
• D. $81$

Answer 1

Answer: (D)

$81$

We know that one of the formula is $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$.

In the given polynomial $4x^2+9y^2+16z^2+12xy+24yz+16xz$, we observe that $a=2x,b=3y$ and $c=4z$, therefore, comparing with the above formula, we get:

$4x^2+9y^2+16z^2+12xy+24yz+16xz=(2x{)}^2+(3y{)}^2+(4z{)}^2+(2\times 2x\times 3y)+(2\times 3y\times 4z)+(2\times 4z\times 2x)$

$=(2x+3y+4z{)}^2$

Let us substitute $x=1,y=1$ and $z=1$ in $(2x+3y+4z{)}^2$ as shown below:

$\left[\right(2\times 1)+(3\times 1)+(4\times 1){]}^2=(2+3+4{)}^2=9^2=81$

Hence, $4x^2+9y^2+16z^2+12xy+24yz+16xz=81$ if $x=1,y=1$ and $z=1$.