Question

Split the middle term 6x^2+7 root 2x+4

Answer 1

Answer:

6x²+7√2x+4

= (3x+2√2)(2x+√2)

Or

= √2(√2x+1)(3x+2√2)

Explanation:

Given quadratic expression:

6x²+7√2x+4

Splitting the middle term, we get

= 6x²+4√2x+3√2x+4

= 6x²+2(2√2)x+3√2x+(√2)(2√2)

= 2x(3x+2√2)+√2(3x+2√2)

= (3x+2√2)(2x+√2)

Or

= (3x+2√2)[√2(√2x+1)]

= √2(√2x+1)(3x+2√2)

Therefore,

6x²+7√2x+4

= (3x+2√2)(2x+√2)

Or

= √2(√2x+1)(3x+2√2)

•••

Answer 2

We split the ‘middle term’ of the $6 x^{2}+7 \sqrt{2 x}+4$, we get the value $\bold{(3 x+2 \sqrt{2})(2 x+\sqrt{2})}$

To find:

Split the ‘middle term’ of the given equation $6 x^{2}+7 \sqrt{2 x}+4$

Solution:

$6 x^{2}+7 \sqrt{2 x}+4$

$=6 x^{2}+4 \sqrt{2 x}+3 \sqrt{2 x}+4$

$=2 \times 3 \times x^{2}+2 \times 2 \sqrt{2 x}+3 \sqrt{2} x+2(2)$

$=2 x(3 x+2 \sqrt{2})+\sqrt{2}(3 x+2 \sqrt{2})$

$=(3 x+2 \sqrt{2})(2 x+\sqrt{2})$

A polynomial is an expression consisting of ‘variables’ and ‘coefficients’ that involves the arithmetic operations like ‘addition’, ‘subtraction’, and ‘non-negative integer exponents’ of variables.