Factorization — grouping and middle term splitting

We need the highest common factor (HCF) of $6x^2$ and $9x$. The HCF of 6 and 9 is 3, and the lowest power of $x$ common to both terms is $x^1$. So the HCF is $3x$.

Divide each term: $6x^2÷3x=2x$ and $9x÷3x=3$. Write these inside the brackets.

The factorized expression is $3x(2x+3)$. Verify by expanding: $3x\times 2x=6x^2$ and $3x\times 3=9x$, giving $6x^2+9x$ ✓

We need to find which option correctly factorizes $6x^2+9x$ by checking which product expands back to the original expression.

Expanding $3(2x^2+3x)=6x^2+9x$. This is technically correct, but the factor inside the bracket can still be factored further since $x$ is common to $2x^2$ and $3x$. This is not fully factorized.

Expanding $6x(x+9)=6x^2+54x$. This gives $54x$, not $9x$, so this is incorrect.

Expanding $9x(x+2)=9x^2+18x$. This gives $9x^2$, not $6x^2$, so this is incorrect.

Only $3x(2x+3)$ expands correctly to $6x^2+9x$ and uses the highest common factor, making it fully factorized.

In the expression $a(x+y)+b(x+y)$, both terms contain the bracket $(x+y)$. This is the common binomial factor.

Treat $(x+y)$ like a single entity. The expression becomes $(x+y)\times (a+b)$, where $a$ and $b$ are what remains after removing $(x+y)$ from each term.

The factorized form is $(x+y)(a+b)$. Verify: $(x+y)(a+b)=ax+bx+ay+by=ax+ay+bx+by$ ✓

We have $a(x+y)+b(x+y)$ and need to factor out the common bracket $(x+y)$. The answer must be a product of two factors.

Expanding $(a+b)(a+b)=a^2+2ab+b^2$, which contains squared terms and doesn't match the original expression. Eliminated.

Expanding $(x+y)(x+y)=x^2+2xy+y^2$, which doesn't contain $a$ or $b$ at all. Eliminated.

Expanding $ab(x+y)=abx+aby$, which has $ab$ as a product in every term, not separate $a$ and $b$ terms. Eliminated.

$(x+y)(a+b)$ expands to $ax+bx+ay+by$, which matches the original expression $ax+ay+bx+by$ ✓