### Video Transcript

There are three parts to this question. Part a) Solve four π₯ squared equals 100.

Any time we see the word βsolve,β we need to calculate the value or values of the unknown β in this case, π₯. In order to solve this equation, we will use the balancing method: whatever we do to one side, we must do to the other.

Our first step here will be to divide both sides of the equation by four. Four π₯ squared divided by four is equal to π₯ squared and 100 divided by four is equal to 25. Our next step is to square root both sides of the equation. The square root of π₯ squared is π₯, as π₯ multiplied by π₯ equals π₯ squared.

When square rooting a number, it is important that we put a positive and negative sign in front of the square root as there will be two possible answers. What number multiplied by itself gives us 25? Well, five multiplied by five is equal to 25. Also, negative five multiplied by negative five is equal to 25, as multiplying two negative numbers gives us a positive answer.

The square root of 25 is equal to positive or negative five. This can be written as one answer or separated into two answers, as shown. The solutions to the equation four π₯ squared equals 100 are π₯ equals five and π₯ equals negative five.

We could check these answers by substituting the numbers back into the initial equation. Five squared is equal to 25 and four multiplied by 25 is equal to 100. Likewise, negative five squared is also equal to 25 and four multiplied by this also gives us an answer of 100.

The second part of the question says the following.

Expand and fully simplify four π₯ minus one multiplied by two π₯ minus three.

In order to expand or multiply out two brackets, we need to multiply everything in the first bracket by everything in the second bracket. One way to do this is using the FOIL method.

Multiplying the first terms four π₯ and two π₯ gives us eight π₯ squared, as four multiplied by two is equal to eight and π₯ multiplied by π₯ is equal to π₯ squared. Multiplying the outside terms four π₯ and negative three gives us negative 12 π₯ as four multiplied by negative three is equal to negative 12.

Next, we need to multiply the inside terms, negative one and two π₯. Multiplying these gives us negative two π₯. Finally, we need to multiply the last terms, negative one and negative three. Negative one multiplied by negative three is positive three. Multiplying two negative numbers together gives us a positive answer.

We were also asked to fully simplify this expression. In order to do this, we need to group or collect the like terms β in this case, negative 12π₯ and negative two π₯. Negative 12π₯ minus two π₯ is equal to negative 14π₯. This means that our simplified expression is eight π₯ squared minus 14π₯ plus three. Four π₯ minus one multiplied by two π₯ minus three is equal to eight π₯ squared minus 14π₯ plus three.

The third and final part of our question says the following.

c) Factorise π₯ squared plus 10π₯ plus 25.

Factorising is the opposite of expanding brackets. In this case, we need to write the quadratic expression π₯ squared plus 10π₯ plus 25 into two brackets. As there is no number in front of the π₯ squared term, we can say that the coefficient of π₯ squared is equal to one. This means that the first term in both brackets will be π₯ as π₯ multiplied by π₯ is equal to π₯ squared.

The second terms in both of the brackets need to have a sum of 10 and a product of 25. The only two pairs of numbers that multiply to give us 25 or have a product of 25 are five and five and one and 25. Which of these pairs of numbers also has a sum or adds up to 10? Five plus five is equal to 10. This means that the second term in both brackets is plus five. The factorisation of π₯ squared plus 10π₯ plus 25 is π₯ plus five multiplied by π₯ plus five.

We could check this answer by using the FOIL method that we used in part b. Multiplying the first terms gives us π₯ squared. The outside terms multiply to give us five π₯ as theyβre the inside terms. And the last terms multiply together to give us 25. Collecting the like terms in the middle gives us π₯ squared plus 10π₯ plus 25, which is what we started with.

Whilst it looks like we have completed this question, there is one more step. As the two brackets are the same, weβre multiplying π₯ plus five by itself. This can be written as π₯ plus five all squared. We can only do this if the two brackets are identical. In this case, the factorisation of π₯ squared plus 10π₯ plus 25 is π₯ plus five all squared.

We now have the answers to all three parts of this question.